Unit 2
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UNIT 2
LIMITS AND CONTINUITY
Structure 2.1
Introduction Objectives
2.2
2.3
2.4 2.5
2.1
Lirnib Algebra of Limits Limits as x += (or - m) One-sided Limits Continuity Definitions and Examples Algebra of Continuous Functions Summary Solutions and Answers
INTRODUCTION
The last unit has helped you in recalling some fundamentals that will be needed in this course. We will now begin the study of calculus, starting with the concept of 'limit'. As you read the later units, you will realise that the seeds of calculus were sown as early as the third century B.C. But it was only in the nineteenth century that a rigorous definition of a limit was given by Weierstrass. Before him, Newton, d' Alembert and CAuchy had a clear idea abort limits, but none of them had given a formal and precise definition. They had depended, more or less, on intuition or geometry. The introduction of limits revolutionised the study of calculus. The cumbersome proofs which were used by the Greek mathematicians have given way to neat, simpler ones. You may zlrbady have an intuitive idea of limits. In Sec. 2 of this unit, we shall give you a precise defi~itionof this concept. This will lead to the study of continuous functions in Sec.3. Most of the functions that you will come across in this course will be continuous. We shall also give you some examples of discontinuous functions.
Objectives After reading this unit you should be able to : calculate the limits of functions whenever they exist, identify points of continuity and discontinuity of a function.
2.2 ,BASIC PROPERTIES OF R In this section we will introduce you to the notion of 'limit'. We start with considering a situation which a lot of us are familiar with, such as train travel. Suppose we are travelling from Delhi to Agrh by a train which will reach Agra at 10.00 a.m. As the time gets closer and closer to 10.00 a.m., the distance of the train from Agra gets closer and closer tdzero (assuming that the train is running on time!). Here, if we consider time as out independent variable, denoted by t and distance as a function of time, say f(t), then we see that f(t) approaches zero as t approaches 10. In this case we say that the limit of f(t) is zero as t tends to 10. Now consider the function f :R +R defined by f(x) = x2+ 1. Let us consider Tables l(a) and 1(b) in which we give the values of f(x) as x takes values nearer and nearer to 1. In Table l(a) we see values of x which are greater than 1. We can also express this by saying that x approaches 1 from the right. Similarly, we can say that x approaches 1 from the left in Table 1(b).
Real Numbers and
Table 1 (a)
Functions
Table 1 @)
We find that, as x gets closer and closer to 1, f(x) gets closer and closer to 2. Alternatively, we express this by saying that as x approaches 1 (or tends to I), the limit of f(x) is 2. Let us now give a precise meaning of 'limit'. Definition 1 Let f be a function defined at all points near p (except possibly at p). Let L be a real number. We say that f approaches the limit L as x approaches p ~ ffor , each real number E > 0, E can find a real number 6 > 0 such that O
' I ~ Idelinitton S of ~ t m was ~t first stated by Karl We~erstrass,around 1850
6 (delta) are Greek letters used to denote real numbers. (& epsilon) and
'+' denotes 'tents to'
x +P
Note that, in the above definition, we take any real number E > 0 and then choose some 6 > 0, s o t h a t L - & < f ( x ) < L + & ,whenever(x-pI<6,thatis,p-6<x
The E - 6 definition does not give us the value Of L. It lust helpts u check whether a given number L is the limit of
Given E > 0. we can choose 6 > 0 such that if we choose x whose distance from p is less than
f(x).
6, then the distance of its image from L must be less than E. The pictures in Fig. 1 may help you absorb the definition.
for each E > 0
3 a 6 > 0 s.t.
O
3
Fig. 1
Remember, the number E is given first and the number 6 is to be produced. An important point to note here is that while taking the limit of f(x) as x -+ p, we are concerned only with the values of f(x) as x takes values closer and closer to p, but not when x = p. For x2 -1 example, consider the function f(x) = -.This function is not defined for x = 1, but is x-1 defined for all other x E R. However, we can still about is limit as x -+ 1. This is because for taking the limit we will have to look at the values of f(x) as x tends to 1, but not when x = 1. Nowlet us take the following examples: Example 1 Consider the function f : R -+ R defined by f(x) = x3. How can we find lim =f(x)? x
-9
0
I f(x) - L I < E
Elements of Differential Calculus
Fig. 2
Look at the graph off in Fig. 2. You will see that when x is small, x3is also small. Asx comes closer and closer to 0, x3 also comes closer and closer to zero. It is reasonable to expect that lirnf(x)=Oasx+ 0. Let us prove that this is what happens. Take any real number E > 0. Then, ( f(x) - 0 I < E w I x3 I < E B ( x ( < &'I3. Therefore, if we choose 6 = &'I3 we get I f(x) - 0 I < E whenever O
terms of (x - a) as much as possible.
Let us now see how to use this rule to calculate the limit in the following examples. x2-1 Example 2 Let us calculate lim x+l x-l We know Aat division by zero is not defined. Thus, the function f(x) =
x2 - 1
x_1 is not defined
at x = 1. But, as we have mentioned earlier, when we calculate the limit as x approaches 1, we do not take the value of the function at x = 1. Now, to obtain lim x+l
x2-1=(x-I)(x+l),sothat,
x2-1
x2 -1 we first note that
x-1
I
2
x - I - lim x_1 =x+lforx+l.Therefore lim - x + 1 (x+ 1). X+l x-1
As x approaches 1, we can intuitively see that this limit approaches 2. To prove that the limit is 2,wefirstwriteflx)-L=x+ 1 -2=x-1,whichisitselfintheformx-a,sincea= 1 inthis case. Let us take any number E > 0. Now, ( ( x + I ) - ~ ( < E w ( x (-<~E Thus, if we choose 6 = E,in our defmiton of limit, we see that ) x - 1 ) < 6 = e a I f ( x ) - L ( = ( x - 1 (<&.Thisshowsthat lim (x+.l)=2.Hence, x
-9
l
x2 -1 lim -= 2. x+l x-1
Example 3 Let us prove that lim (x2+ 4) = 13. x+3
Thatis,weshallprovethat VE>O. 36>Osuchthat)x2+4- 1 3 ) < ~ w h e n v e r ( x - 3 ) < 6 .
Real N u m b e r s and Functions
Nowwewrite)x2- 9 / i n t r m s o f ( x - 3 1 : 1x2-9I=lx+3)(x-31 Thus, apart from 1 x - 3 1, we have a factor, namely I x + 3 ( of [x2- 91. To decide the limits of 1 x + 3 1, let us put a restriction on 6. Remember, we have to choose 6. So let us say we choose a 6 5 1. What does this imply? Ix-3)<6 3(x-3)<133-1<x<3+1
= + 2 < ~ < 4 3 5 < ~ + 3 < 7 . R e c a I l S e c . 4 , U n1.i t Thus, we have / x2- 9 ( < 7 1 x - 3 / < E.Now when will this be true? It will be true when i x - 3 1 w7. So this d 7 is the value of 6 we were looking for. But we have already chosen 6 < 1. This means that given E > 0, the 6 9 e choose should satisfy 6 5 1 and also 6 5 ~ 2 7 . In other words, 6 = min { 1, d7}, should serve our purpose. Let us verify this : ! x - 3 ) < 6 * I x - 3 1 < 1 a n d ~ x - 3( < d 7 * I x 2 - 9 1 = ) x + 3 l l x - 3 ) < 7 . d 7 = ~ .
Remark 1 : Iff is a constant function on R, that is, if f(x) = k
x E RI where k is some fixed real
number, then ,lim ,f(x) = k. ~ o w b l e a s try e the following exercises. E 1) Show that 1 lim - = 1 a) x,, x
b)
x3 - 1
lim -= 3. x+l x-1
Before we go further, let us ask, 'Can a function f(x) tend to two different limits as x tends to p '? The answer is NO, as you can see from the following :
Elements o f Differential calculus
*Theorem 1 If lim fix) = L and lim fix) = M, then L = M. Proof :Suppose L # M, then ( L - M I > 0. Since lim f(x) = L. If we take E = I L - M I 'J -'P 2 then 3 6, > 0 such that Similarly, since lim f(x) = M, 3
> 0 such that
P 'x-
Ifwechoosie6=rnin~6,,6,1,then6>OandIx-p(<6willmeanthat(x-p)<6,and IX--PI<~~. In this case we will have both ( f(x) - L ( < E,as well as, I f(x) - M I < e.
Sothat(L-M)=)L-f(x)+f(x)-M(lIf(x)-LI+(f(x)-MI<e+~=2&=IL-M(. That is, we get I L - M 1 < ( L - M 1, which is a contradiction. Therefore, our supposition is wrong, Hence L = M. We now staite and prove a theorem whose usefulness will be clear to you in Unit 4. Theorem 2 Let f, g, and h be functions defined on an interval I containing a, except possibly at a, Suppose
i)
lirn f (x) = L = lim h (x)
x+a
x+a
Then lirn g (x) exist and is equal to L. x-+ a
Proof: By the definition of limit, given E > 0 , 3 6, > 0 and 6, > 0 such that (f(x)-LI<&forOI~-aI<6,and Ih(x)-L)<eforO(x-aI<6,. 'Let 6 = min ( 4 , 6,). Then, O<(x-a1<6* If(x)-L(<&andIh(x)-LICE 3L-eIf(x)
Theorem 2 is also called the sandwich theorem (or the squeeze theorem), because g is being smdwiched between f and h. Let us see how this theorem can be used.
Example 4 Given that I f(x)- 1 ( 5 3(x +
t/ x E R, can we calculate
lim f (x) ?
x+-1
We know that-3(x + S f(x) -1 S 3 (x + v x. This means that -3(x + + 1 5 f(x) 5 3(x + 1)2+ 1 5 t~ x. Using the sandwich theorem and the fact that lirn [-3 (x+
x+-1
+ 1]= 1 = lirn [3(x+ I-'-.?
+ I], weget lirn f(x)= 1 x-'-
I
In the next section we will look at the limits of the sum, product and quotient of functions.
Algebra of limits
2.2.1
Now that you are familiar with limits. Let us state some basic properties of limits. (Their proofs are beyond and scope of this course.) 'Theorem3 Let f and g be two h c t i o n s such that lim f(x) and lim g(x) exist. Then P'-X
i)
P'-X
lim[f(x)+g(x)]=limf(x)+limg(x) P'-X
ii)
P'-X
lim [f(x) g (x)] = X-'P
5) lim X+P
[:yp
Sum rule
P 'x-
f(x)
]
lim g(x)
1 = I' ,provided lim g (x) gdx) litn g (x) x-rp
P'-X
Productrule t0
Reciprocal rule
We can easily prove two more rules in addition to the three rules given in Theorem 3. These are : *-' lirn k = k
iv)
Constant function rule
X+P
v)
limx=p
Identity function rule
x +P
We shall only indicate the method of proving iv) and v): iv) Here I f(x) - L I = 1 k - k I = 0 < E, whatever be the value of 8. *
Using the properties that we have just stated, we will calculate the limit in the following example. Example 5 Let us evaluate lim x+2
3x2 + 4x
2x + 1
NOW lim 2~ + 1 = lirn 2x + lim 1 by using i) x+2
x+2
lirn 2 lirn x + lirn 1 by using ii)
=
x+2
I
P
:.
x+2
x+2
= 2 x 2 + 1=5+Obyusingiv)andv) We can use (iii) of Theorem 3. Then the required limit is
lirn 3 x 2 + lirn 4x
lirn (3x2 + 4x)
x+2
lim ( 2 +~1)
!
x+2
x+2
-
x+2
lim 2 x + liml
x+2
x+2
b~u~ingi)
x+2
lirn 3 lim x lirn x + lim 4 lirn x
x+2
-
x+2
x+2
x+2
lim 2 lirn x + lim 1
x+2
x+2
x+2
by using ii)
x+2
3 x 2 x 2 + 4 x 2 = -20 - -4 2x2+1 5 You can similarly calculate the limits in the following exercises. -
E
E E3) t
3
E2) Show that lim - = 3 X+l
X
Calculate lirn x + l 2x +
(A)
Real Numbers and Functions
Elements o f Differential Calculus
2.2.2 Limits as x
-+
-
(or - -) 1
Take a look at the graph of the hnction f(x) = - ,x > 0 in Fig. 3. This is a decreasing function X
of x. In fact, we see fromFig. 3 that f(x) comes closer and closer to zero as x gets larger and larger. This situation is similar to the one where we have a hnction g(x) getting closer and closer to a value L as x comes nearer and nearer to some number p, that is when lim g(x) = L. X-'P
The only difference is that in the case of f(x), x is not approaching any finite value, and is just becoming larger and larger. We express this by saying that f(x) + 0 as x + w ,or lim f(x) = 0. X--W
Note that, = is not a real number. We write x + w merely to indicate that x becomes larger and larger. We now formalise this discussion in the following definition. Fig. 3
Definition 2 A function f is said to tend to a limit L as x tends to w if, for each E > 0 it is possible to choose K > 0 such that I f(x) - L I < E whenever x > K. In this case, as x gets larger and larger, f(x) gets nearer and nearer to L. We now give another example of this situation. Example 6 Let f be defined by setting f(x) = 1/x2for all x E R \ (0). Here f is defined for all real values of x other than zero. Let us substitute larger and larger values of x in f(x) = 1/x2and see what happens (see Table 2). Table 2
We see that as x becomes larger and larger, f (x) comes closer and closer to ie:o: Now, let us choose any E > 0. If x > l! J;, then 1/x2< E. Therefore, by choosing K = 11 &, we find that x >K
i f(x) / < E. Thus, lim
f(x) = 0.
X-SQ,
Fig. 4 gives us a graphic idea of how this function behaves as x + w.
AY
Fig. 4
Sometimes we also need to study the behaviour of a function f(x), as x takes smaller and smaller negative values. This can be examined by the following definition. Definition 3 A function f is said to tend to a limit L as x + - if, for each E > 0,it is possible to choose K > 0, such that / f(x) - L I < E whenever x < - K. The following example will help you in understanding this idea. Example 7 Consider the function f : R + R defined by
The graph o f f is as shown in Fig. 5.
LImltr and Continuity
What happens to f(x) as x takes smaller and smaller negative values? Let us make a table (Table 3) to get some idea. Table 3
We see that x takes smaller and smaller negative values, fix) comes closer and closer to zero. In fact 141 + x2)< E whenever 1 + x2> I/&,that is, whenever x2> (I/&)- 1, that is, whenever either
x < - k 3 I f(x) j C E. Consequently, lim fix) = 0. X+-@
In the above example we also find that lim f (x) = 0. x-bm
Let us see how lim f(x) can be interpreted geometrically.
-
x+m
In the above example, we have the function f(x) = 141 + x2),and as x -+ m, or x -+ - m, f(x) -+0. From Fig. 5 you can see that, as x -+ or x -+ - m ,the curve y = f(x) comes nearer and nearer the straight line y = 0, which is the x-axis. Similarly, if we say that lim g(x) = L, then it means that,as x -,m the curve y = g (x) comes x+m
closer and closer to the straight line y = L. x - 1. Example 8Let us show that lim -- X-'m (1 + x2)
I
Now, 1 x2/(1+ x2)- 1 1 = 141 + x2). In the previous example we have shown that 1 1/(1+ x2) ) < E for x > K, where K = 1 I/€- 1.' 1 Thus, given E > 0, we choose K = / 1 / -~1 JIJZ, so that
We show this geometrically inFig. 6. 4y
1
Fig. 6
-
Elements o f Differential Calculus
You must be wondering if all the properties given in Theorem 3 also hold when we take limits as x + . Yes, they do. You can use them to solve this exercise.
E
E4) Shos that
a)
lirn l/x=O x-bm
b)
lim (l/x+3/x2+5)=5 X-bco
Sometimes we cannot use Theorem 2 directly, as is clear in the following example, Letus see how to overcome this problem. 3x + 1 Example 9 Suppose, we want to find xlirn -, 2x + 5
'
-
We cannot apply Theorem 3 directly since the limits of the numerator and the denominator, as x + ,cannot be found. Instead, we rewrite the quotient by multiplying the numerator and denominator by llx. for x#O. 3x+l - 3+(l/x) Then, 2 x - 2 + (5 I x) , for x # 0. Now we use Theorem 3 and the fact that lirn llx = 0 (see # 4 a), to get x-bm
3 + (1 / x) 3x + 1 lim = lim x-+m2x+5 x - + m 2 + ( 5 / x )
-
lim ( 3 + l / x ) x-+m
- lirn ( 2 + 5 / x ) x-+m
E
3+0
- -=-
2+0
-
2
By now you knust be used to the various defmitions of limits, so can try this exercise E5) a) If for some E > 0,and for ever K, 3 x >K s.t. I f(x) - L 1 > E, what will you infer? b) If lirn f(x) ;f L, how can you express it in the E - 6 form? x -tP
Real Numbers and Functions
We end this section with the following important remark. Remark 2 In case we have to show that a function f does not tend to a limit L as x approaches p, we shall have to negate the definition of limit (also see E5(b)). Let us see what this means. Suppose we want to prove that lim f(x) ;t L. Then, we should find some E > 0 such that for every 6 > 0, there is some x E ]p Y8for which I (x) - L I > 6. Through our next example we shall illustrate the negation of the definition*ofthe limit of f(x) as x +00. Example 10 To show that lirn llx # I, we have to find some E 0 such that for any K (howsoever large) we can alGzys find an x > K such that ( I/x - 1 I > E. 114. This clearly show that lim l/x # 1. X+(D
2.2.3
One-sided Limits
If we consider the graph of the function f(x) = [XI,shown in Fig. 7, we see that f(x) does not see to approach any fixed value as x approaches 2. But from the graph we cansay that if x approaches 2 from the left then f(x) seems to tend to 1. At the same time, if x approaches 2 from the right, then f(x) seems to tend to 2. This means that the limit o f f cxists if x approaches 2 from only onc side (left or right ) at a time. This example suggests that we in!roduce the idea of a one-sided limit. Definition 4 Let f be a function defined for all x in the interval ]p, q[. f is said to approach a limit L as x approaches p from right if, given any E > 0, there exist a number 6 > 0 such that p<x
Similarly, the function f : )a, p[ + R is said to approach a limit L as x approaches p f r o p the left i f , g i v e n a n y ~ > 0 , 3 6~ O s u c h t h a t p - - 6 < x < p ~ I f ( x ) - L I < & . This limit is denoted by lirn f (x). x+p-
Note that in computing these limits the values of f(x) for x lying on only one side of p are taken ~ n t oaccount. Let uf apply this definition to the function f(x) = [ x 1, we know that for x E [l, 2[, [x] = 1. That is, [x] is a constant function on [I, 2[. Hence lim [x] = 1. Arguing similarly, we find that since [x] x+2-
=:
2 for all x E [2,3[, [XIis, again, aconstant function on [2,3[, and lim [x] = 2. x+2+
Let us improve our understanding of the definition of one-sided limits by lookiilg at some more examples. u
Example 11 Let f be defined on R by setting
We will show that lirn f(x) equals -1. x 4 -
When x < 0 , )x 1 = - x, and therefore, f(x) = (- x)lx = -1. In order to show that lim f(x) exists x-0-
and equals -1, we have to start with any E > 0 and then find a 6 > 0 such that, if
-6<x 0 will work. Therefore, whateGer 6 0 we may choose, if - 6 < x < 0,then 1 f(x) - (-1) I = 0 < E. Hence lim f (x) x 4 -
Example 12 f is a function defined on R by setting f(x)= x - [x], for all x E R. Let us examine whether lim f(x) exists. x+1-
Recall (Unit 1) that this function is given by f(x) = x, if 0 5 x < 1. f(x) = x - 1 if 1 5 x < 2, and, in general
= -1.
Fig. 7
Elements of Differential Calculus
Fig. 8
Since f(x) = x for values of x less than 1 but closc to 1, it is reasonabie to expect that lim f(x) = 1. Let us prove this by taking any E : O and choosing 6 = min / 1, E I. We f h d x-1-
1-6<x
I=Ix-l ) < 6 < & .
Therefore. l i n ~ f(x) = 1. X-+l
Proceeding exactly as above, the noting that f(x) = x - 1 if 1 5 x < 2, we can similarly prove that f(x) =O.
E
5 6 ) Prove that
a)
lim x [x] -4 x-3-
b)
lim -1x1 -=I
n+Oi
X
Before going further, !et us see how the concepts of one-sided limit and limit are comected.
Real
Theorem 4 'The following statements are equivalent. lirn qx)
i)
exists
X+P
lirn f ( X )
ii)
and l i n ~ exist and are equal. x-+p-
Proof :'To show that i) and ii) are equivalent, we have to show that i) 3 ii) and ii) s i). We first prove that i) 3 ii). For this we assume that lirn f(x) = L. Then given *E > 0.
"36>@suchthat]f(x)-L)<~forO
X-+P
lirn f ( x ) = L z lim f(x). x+p-
x+p+
We now prove the converse, that is, ii) 3 i). For this, we assume that
lirn f ( x ) :- linl f(x) = L. Then, givcii E > 0,36, ,6, > 0. x+p-
x+pt
such that if(x)-Li<~forp-6,<x
Hence, lirn fjx) =L. X+P
Thus, we have shown that i) =, ii) and ii) 3 i), proving that they are equivalent. From Theorem 4, we can conclude that if lim qx) exists, then lirn f(x) and Aim f(x) x-' P x-tp* x-tpalso exist and further. lim f(x)= lirn f(x)= lirn f(x) x
P
x-tp+
-+
x4p-
Remark 3 If you apply Theorem 4 to the function f(x) = x - [x] (see Example 12), you will see that lim {x - [x]} does not exist as lirn (x - [x]) # lim {x- [x]} , x-1
x+1+
x+l-
We shall use this concept of one-sided limits to define continuous functions in the nest section.
2.3 CONTINUITY A continuous process is one that goes on smoothly without any abrupt change. Continuity of a function can also be interpreted in a similar way. Look at Fig. 9. The grapi? of the function f in Fig. 9 (a) has an abrupt cut at the point x = 3, whereas the graph of the function g in Fig. 9 (b) proceeds smoothly. We say that the function g is continuous, while f is not.
~4
y
4
(a)
(b) Fig. 9 : (a) Graph of f (b) Graph of g
Continuous functions play a very important role in calculus. As you proceed, you will be able to see that many theorems which we have stated in this course are true only for continuous functions. You will also see that continuity is a necessary condition for the derivability of a function, and that it is a sufficient condition for the integrability of a function. Bilt let us give a precise meaning to "a continuous function'' now.
umbers
and Functions
E i e n ~ e n t sof Differential Calculus
Definitions and Examples
2.3.1
In this sectLon we shall give you the definition and some examples of a continuous function. We shall also give you a short list of conditions which a function must satisfy in order to be continuous at a point.
Definition 5 Let f be a function defined on a domain D, and let r be a positive real number such that the interval ]p - r, p + r[ c D. f is said to be continuous at x = p if lim f(x) = f(p). X+P
,
By the definition of lim~tthls means that f is continuous at p is given E > 0 , 3 6 > 0 such that I fix)
- f(p) ) < F whenever 1 x - p j < 6. To clarify this concept let us look at an example
'
Example 13 L.et us check the continuity of the function
-
f : L R 4 W such that f(x) x at the point x = 0. Now, f(0) = 0. Thus we want to know if lirn f (x) = 0. x+O
This is true because given E > 0,,we can choose 6 = E and verify that ( x ( < 6 3 I f (x) I < E. Thus f is aontinuous at x = 0.
Remark 4 f is cont~nuousat x = p provided the following two criteria are met : i)
.
lim f(x)
x+p
exists.
Fig. 10 : (a) Graph of f
(b) G r a ~ hof g
Fig. 1U ~lioiwstwo discontinuous functions f and g. Criterion i) IS not inet by f, whereas g fails to meet criterion ii). If you read Remark 3 again, you will find that f(x) = x - [x] is not continuous as x = 1. But we have see that we can calculate one-sided limits of f(x) = x - [x] at x = 1. This leads us to the following definition.
Dennition 6 A function f : lp, q[ R is said to be continuous from the right at x == p ~f lirn f(k) = f (p). We say that f is continuous from the left at q if lirn f(x) = f (q). x+pt
x+q-
Thus, f(x) = x - [xl is continuous from the right but not from the left at x = 1 since lim f ( x ) x-1-
it f(1) and
lirn f(x)- f(1) =0. x+l'
E E 7) Give E - 6 definition of continuity at a point from the right as well as from the le'ft.
E
E 8) Show that function f: ]p - r, ,p + r[ -+ R is continuous at x = p if and only i f f is continuous from the right as well as from the left at x p (Use Theorem 4).
Real Numbers and Functions
;-
Now that you know how to test the continuity of a function at a point, let us go a step further, and define continuity of a function on a set.
Definition 7 A function f defined on a domain D is said to be continuous on D, if it is continuous at every point of D. Let us see some more examples. Example 14 Let f(x) = xn for all x E R and any n E Zt. Show that f(x) is continuous at x = p for all p E R. We know that lirn x = p for any p E R. Then, by the product rule in Theorem 3, we get
z' is the set of positive integers
X+F
lirn xn = ( lirn x) ( lirn x) ..........( lim x) X+P
-
x+p
x+p
(n times)
X+P
pp ............. p (n times) =p". Therefore, lirn f(x) exists and equals f(p). Hence f is x+p. continuous at x = p. Since p was any arbitrary number in R, we can say that f is continuous on R. Remark 5 Using Example 14 and Theorem 3, we can also prove that polynomial a, + alx + ............ + allx", where a,, a, ..............an E R, is continuous on R, that is, l i m ( a o + a l x + ................. + a,,xn)= a, + alp + ..............+ anpnfor all p E R. -
X-tP
Example 15 The greatest integer function f : R -+ R : f(x) = [ x ] is cliscontinuous at x = 2. To prove this we recall our discussion in Sec. 2 in which we have proved that lim f(x) = 1 x+2-
and lirn f(x) = 2. Thus, since these two limits are not equal lirn f(x) does not exist. x+2+
-
x+2
Therefore, f is not continuous at x = 2 because the first criterion laid down in Remark 4 is not met. Example 16 Let f(x) = / x / for all x E R. This f is continuous at x = 0. Here f(x) = x, if x 2 0, and f(x) = - x if x < 0. You can show that lirn f(x)= lim x=O=f(O)and
C
x+o*
x+O'
lim f(x) = lirn (-x)
x+o
f
1
x+o-
= 0 = f(0). Thus
lirn f(x) exists and equals f(0). Hence f is continuous
x+o
atx=O. Note : f is also continuous at every other point of R. (Check this statement). x2 - 1
Example 17 Suppose we want to find whether f(x) = -is continuous at x = 0. x-1 In Fig. 11 you see the graph o f f . It is the line y = x + 1 except for the point (1,2) Y'
Fig. 11
We can write f(x) all x 1.'
=
x
+ 1 for
Elements of Dlfferentlal
f(o)=(02- 1)/(0- I ) = 1 and limf(x)= lim (x2-l)/(x-1)=
Calculus
x-ro
1
x-to
lim ( x i l ) = l
x+o
lim f(x) = f(O), so that f is continuous at x = 0.
x+O
The exponential function f(x) =ex and the logarithmic h c t i o n f(x) = In x are continuous functions. You can check this by looking at their graphs in Unit 1. Similarly,x +sinx and x + cosx are continuous, x + tanx is continuous in ] - d 2 , n;2[. This fact is quite obvious from the graphs of these functions (We have given their graphs in Unit 1). We shall not attempt a rigorous proof of their continuity here. Caution :Checking the continuity of a function from the smoothness of its graph is not a fool-proof method. If you look at the graph (Fig. 12) ofthe h c t i o n x +x sin (llx), you will find that it has no breaks in the neighbourhood of x = 0. But this function is not continuous. Observe that the graph oscillates wildly near zero.
t
. Fig.
E
12
E9) Show that the function f : R + R given by f(x) = 1!(x2 - 9) is continuous at all points of Rexceptatx=3andx=-3.
Now that we know how to check whether a function is continuous or not, let us go further, and talk about the continuity of some combinations of functions.
2.3.2 Algebra of Continuous Functions Let f and g be functions defined and continuous on a common domain D E R, and let k be any real number. In Unit 1, we defied the functions f + g, fgl Ug (provided g(x) z 0 any where in D), kif and I f I. The following theorem tells us about the continuity of these functions.
Theorem 5 Let f and g be functions defined and continuous on a common domain D, and let k be any real number. The functions f + g, kf, 1 f I and fg are all continuous on D. If g(x) # 0 anywhere in D, then the function flg is also continuous on D. We shall not prove this theorem here. In Unit 1, you have studied the important concept of composite functions. In Theorem 6, we will talk about the continuity of the composite of two continuous functions. Here again, we shall state the theorem without giving proof as it is beyond the level of this course.
Theorem 6 Let f : D, + D2and g :D2 + D, be continuous on their domains. Then gof is continuous on Dl, (Dl, D2,D3c R). Example 18 To prove that f : R +R : f (x) = (x2+ is continuous at x = 0, we consider the functions g : R +R : g(x) = x3 and h : R +R :h(x) = x2+ 1. You can check that f(x) = goh (x). Further, by Remark 5, h is continuous on R, and g is also continuous on R. Thus, goh = f is continuous on R. Let us see if the converse of the above theorems are true. For example, iff and g are defined on an interval [a, b] and iff + g is continuous on [a, b], does that mean that f and g are continuous on [a, b] ? No. Consider the functions f and g over the interval [0, 11 given by
Then neither f nor g is continuous at x = 112. (Why?) But (f + g) (x) = 1 v x E [O, 11. Therefore, f + g is continuous on [0, 11. Now, if ( f I is continuous at a point p, must f also be continuous at p? Again, the answer is No.Take, for example, the function f : R +R givenby f(x) =
-1, 1,
forx1O for > 0
Then I f(x) 1 = 1 in Rand hence I f 1 is continuous. But f is not continuous at x = 0 (Why?) 1,
E10) Iff : R +R is defined by f (x) = f(x) = is f continuous at a ) x = 1
C
b) x = -3/2?
-1.
i f x ~ Z ifxeZ
Limits and Continuity
,
E l e m e ~ ~ tosf Differential Calculus
One again, we won't prove this theorem here. But try to understand its statement because we shall be using it in subsequent units. Theorem 7 (Intermediate Value Theorem) Let f be continuous on the closed interval [a, b]. Suppose c ia a real number lying between f(a) and f(b). (That is, f(a) < c < f(b) or f(a) > c > (b?). Then there exists some x, E ]a, b[, such that f(xo)= c. How can we interpret this geometrically? We have already seen that the graph of a continuous function is smooth. It does not have any breaks or jumps. This theorem says that, if the points (a, f(a)) and (b, f(b)) lie on two opposite sides of a lide y = c (see Fig. 13), then the graph o f f must cross the line y = c.
Fig. 13
Note that this theorem guarantees only the existence of the number x,. It does not tell us how to find it. Another thing to note is that this x, need not be unique. That brings us to the end of this unit.
SUMMARY
2.4
Wh end this untt by summarising what we have covered in it.
.
The limit of a function f a t a point p of its domain is L is given E > 0 , 3 6 > 0, such that )f(x)-L I <&whenever1x-pI<6. One-sided limits
3.
p,,
1.
lim f(x) exists if and only if lim
x+p+
4.
f (x)and? -,, lim
f(x) both exist and are equal.
A fuqction f is continuous at a point x = p if lim f (x) = f (P)
,,
5.
6.
If the function f and g are continuous on D, then so are the functions f + g, fg, ( f 1, kf (where k E R) flg (where g(x) # 0 in D) The Intermediate Value Theorem : Iff is continuous on [a, b] and if f(a) < c < f(b) (or f(a) > c > f(b)), then 3x0€ ]a, b[suchthat f(xo)=c. 4
2.5
QOLUTIONS AND ANSWERS
E 1) a) Given any E > 0,if we choose 6 = min {d2,1/2) then I ~ - l I < 6 < 1 / 2 - x>1/2.
Real Numbers and Functions
Hence lim l/x = 1 x+l
Given E > 0, ifwe choose 6
= min
1 (217) E, 1/21,then
Ix-I(<1/2*~<3/2*~+2<7/2and
x3 - 1 Hence, lim -= 3 x+l x - 1 lim 3 1 E2) lirn 3 / x = + = 3 / 1 = 3 x+l hm x x+l
E3) lirn 2 x + 5 x+l
f\ I + x ) =21im x + 15+lirnl i mx2x 2 x2
X+I
E4) a) Given e > 0 if we choose K = lle, then x > K - I l/x-OI=) l / x I < l / K = ~ Thus, lirn l/x=O x+m
b) Given E > 0, if we choose K = 11& ,then x > K * 1 1/x2-OI=( l / x 2 ( i : 1 / K 2 = ~ . Hence, lirn 1/x2=O x+m
Now, lirn (l/x +.3/x2+ 5) x+m
=
lirn 1/x+3 lirn 1/x2+ lirn 5 = 0 + 3 x 0 + 5 = 5 x+m
x+m
E5) a)
x+m
lirn f(x)#L x+m
lirn X-[XI=
x+3-
b)
lirn X+O+
;;y-
x-2=l
I X \ / X J= ~:J+' X / X = ~ , I X I = X ~ O ~ X > O .
- x+olim -(x2+2)=-2 E 7) f is continuous from the right at x = p if v E > 0 there exists a 6 > 0 s.t. p<~ 0 there exists a 6 > 0 s.t. p - 6 < x < p * If(x)-f(p))<~. E 8) f is continuous at x = p
* !!4x) = f(p)
lim f(x) = f (p) and x+plim * ,+,+
f (x) = f(p) by Theorem 4.
f is continuous from right and from left at x = p. Iff is continuous from right and left,
Elements of Differential Calculus .
lim 3
!$ fix) exists and = qp)
3f
is continuous at p,
E9) lim f(x) = x-+a
lim x2 - 9
--
1
a2 - 9
= f (a) for all a except a = 3 f 3
X+l
Hence f is continuous at all points except at f 3, f is not defined at 23. E10) a) f is not continuous at x = 1. For E = 1 and any 8 > 0, if x is any non-integer E ] 1-8,1+8[,then ( f l ~ ) - f ( l ) ( = I - l - lI = ~ > E . b) f is continuous at -312. Since given E > 0, if we choose 8 < 112, then
Ix-(-312)1< 1 1 2 3 - 2 < x < - 1 3 x 4 Zandhence )f(x)-f(-312) /=O<E.
LIMITS AND CONTINUITY
Structure 2.1
Introduction Objectives
2.2
2.3
2.4 2.5
2.1
Lirnib Algebra of Limits Limits as x += (or - m) One-sided Limits Continuity Definitions and Examples Algebra of Continuous Functions Summary Solutions and Answers
INTRODUCTION
The last unit has helped you in recalling some fundamentals that will be needed in this course. We will now begin the study of calculus, starting with the concept of 'limit'. As you read the later units, you will realise that the seeds of calculus were sown as early as the third century B.C. But it was only in the nineteenth century that a rigorous definition of a limit was given by Weierstrass. Before him, Newton, d' Alembert and CAuchy had a clear idea abort limits, but none of them had given a formal and precise definition. They had depended, more or less, on intuition or geometry. The introduction of limits revolutionised the study of calculus. The cumbersome proofs which were used by the Greek mathematicians have given way to neat, simpler ones. You may zlrbady have an intuitive idea of limits. In Sec. 2 of this unit, we shall give you a precise defi~itionof this concept. This will lead to the study of continuous functions in Sec.3. Most of the functions that you will come across in this course will be continuous. We shall also give you some examples of discontinuous functions.
Objectives After reading this unit you should be able to : calculate the limits of functions whenever they exist, identify points of continuity and discontinuity of a function.
2.2 ,BASIC PROPERTIES OF R In this section we will introduce you to the notion of 'limit'. We start with considering a situation which a lot of us are familiar with, such as train travel. Suppose we are travelling from Delhi to Agrh by a train which will reach Agra at 10.00 a.m. As the time gets closer and closer to 10.00 a.m., the distance of the train from Agra gets closer and closer tdzero (assuming that the train is running on time!). Here, if we consider time as out independent variable, denoted by t and distance as a function of time, say f(t), then we see that f(t) approaches zero as t approaches 10. In this case we say that the limit of f(t) is zero as t tends to 10. Now consider the function f :R +R defined by f(x) = x2+ 1. Let us consider Tables l(a) and 1(b) in which we give the values of f(x) as x takes values nearer and nearer to 1. In Table l(a) we see values of x which are greater than 1. We can also express this by saying that x approaches 1 from the right. Similarly, we can say that x approaches 1 from the left in Table 1(b).
Real Numbers and
Table 1 (a)
Functions
Table 1 @)
We find that, as x gets closer and closer to 1, f(x) gets closer and closer to 2. Alternatively, we express this by saying that as x approaches 1 (or tends to I), the limit of f(x) is 2. Let us now give a precise meaning of 'limit'. Definition 1 Let f be a function defined at all points near p (except possibly at p). Let L be a real number. We say that f approaches the limit L as x approaches p ~ ffor , each real number E > 0, E can find a real number 6 > 0 such that O
' I ~ Idelinitton S of ~ t m was ~t first stated by Karl We~erstrass,around 1850
6 (delta) are Greek letters used to denote real numbers. (& epsilon) and
'+' denotes 'tents to'
x +P
Note that, in the above definition, we take any real number E > 0 and then choose some 6 > 0, s o t h a t L - & < f ( x ) < L + & ,whenever(x-pI<6,thatis,p-6<x
The E - 6 definition does not give us the value Of L. It lust helpts u check whether a given number L is the limit of
Given E > 0. we can choose 6 > 0 such that if we choose x whose distance from p is less than
f(x).
6, then the distance of its image from L must be less than E. The pictures in Fig. 1 may help you absorb the definition.
for each E > 0
3 a 6 > 0 s.t.
O
3
Fig. 1
Remember, the number E is given first and the number 6 is to be produced. An important point to note here is that while taking the limit of f(x) as x -+ p, we are concerned only with the values of f(x) as x takes values closer and closer to p, but not when x = p. For x2 -1 example, consider the function f(x) = -.This function is not defined for x = 1, but is x-1 defined for all other x E R. However, we can still about is limit as x -+ 1. This is because for taking the limit we will have to look at the values of f(x) as x tends to 1, but not when x = 1. Nowlet us take the following examples: Example 1 Consider the function f : R -+ R defined by f(x) = x3. How can we find lim =f(x)? x
-9
0
I f(x) - L I < E
Elements of Differential Calculus
Fig. 2
Look at the graph off in Fig. 2. You will see that when x is small, x3is also small. Asx comes closer and closer to 0, x3 also comes closer and closer to zero. It is reasonable to expect that lirnf(x)=Oasx+ 0. Let us prove that this is what happens. Take any real number E > 0. Then, ( f(x) - 0 I < E w I x3 I < E B ( x ( < &'I3. Therefore, if we choose 6 = &'I3 we get I f(x) - 0 I < E whenever O
terms of (x - a) as much as possible.
Let us now see how to use this rule to calculate the limit in the following examples. x2-1 Example 2 Let us calculate lim x+l x-l We know Aat division by zero is not defined. Thus, the function f(x) =
x2 - 1
x_1 is not defined
at x = 1. But, as we have mentioned earlier, when we calculate the limit as x approaches 1, we do not take the value of the function at x = 1. Now, to obtain lim x+l
x2-1=(x-I)(x+l),sothat,
x2-1
x2 -1 we first note that
x-1
I
2
x - I - lim x_1 =x+lforx+l.Therefore lim - x + 1 (x+ 1). X+l x-1
As x approaches 1, we can intuitively see that this limit approaches 2. To prove that the limit is 2,wefirstwriteflx)-L=x+ 1 -2=x-1,whichisitselfintheformx-a,sincea= 1 inthis case. Let us take any number E > 0. Now, ( ( x + I ) - ~ ( < E w ( x (-<~E Thus, if we choose 6 = E,in our defmiton of limit, we see that ) x - 1 ) < 6 = e a I f ( x ) - L ( = ( x - 1 (<&.Thisshowsthat lim (x+.l)=2.Hence, x
-9
l
x2 -1 lim -= 2. x+l x-1
Example 3 Let us prove that lim (x2+ 4) = 13. x+3
Thatis,weshallprovethat VE>O. 36>Osuchthat)x2+4- 1 3 ) < ~ w h e n v e r ( x - 3 ) < 6 .
Real N u m b e r s and Functions
Nowwewrite)x2- 9 / i n t r m s o f ( x - 3 1 : 1x2-9I=lx+3)(x-31 Thus, apart from 1 x - 3 1, we have a factor, namely I x + 3 ( of [x2- 91. To decide the limits of 1 x + 3 1, let us put a restriction on 6. Remember, we have to choose 6. So let us say we choose a 6 5 1. What does this imply? Ix-3)<6 3(x-3)<133-1<x<3+1
= + 2 < ~ < 4 3 5 < ~ + 3 < 7 . R e c a I l S e c . 4 , U n1.i t Thus, we have / x2- 9 ( < 7 1 x - 3 / < E.Now when will this be true? It will be true when i x - 3 1 w7. So this d 7 is the value of 6 we were looking for. But we have already chosen 6 < 1. This means that given E > 0, the 6 9 e choose should satisfy 6 5 1 and also 6 5 ~ 2 7 . In other words, 6 = min { 1, d7}, should serve our purpose. Let us verify this : ! x - 3 ) < 6 * I x - 3 1 < 1 a n d ~ x - 3( < d 7 * I x 2 - 9 1 = ) x + 3 l l x - 3 ) < 7 . d 7 = ~ .
Remark 1 : Iff is a constant function on R, that is, if f(x) = k
x E RI where k is some fixed real
number, then ,lim ,f(x) = k. ~ o w b l e a s try e the following exercises. E 1) Show that 1 lim - = 1 a) x,, x
b)
x3 - 1
lim -= 3. x+l x-1
Before we go further, let us ask, 'Can a function f(x) tend to two different limits as x tends to p '? The answer is NO, as you can see from the following :
Elements o f Differential calculus
*Theorem 1 If lim fix) = L and lim fix) = M, then L = M. Proof :Suppose L # M, then ( L - M I > 0. Since lim f(x) = L. If we take E = I L - M I 'J -'P 2 then 3 6, > 0 such that Similarly, since lim f(x) = M, 3
> 0 such that
P 'x-
Ifwechoosie6=rnin~6,,6,1,then6>OandIx-p(<6willmeanthat(x-p)<6,and IX--PI<~~. In this case we will have both ( f(x) - L ( < E,as well as, I f(x) - M I < e.
Sothat(L-M)=)L-f(x)+f(x)-M(lIf(x)-LI+(f(x)-MI<e+~=2&=IL-M(. That is, we get I L - M 1 < ( L - M 1, which is a contradiction. Therefore, our supposition is wrong, Hence L = M. We now staite and prove a theorem whose usefulness will be clear to you in Unit 4. Theorem 2 Let f, g, and h be functions defined on an interval I containing a, except possibly at a, Suppose
i)
lirn f (x) = L = lim h (x)
x+a
x+a
Then lirn g (x) exist and is equal to L. x-+ a
Proof: By the definition of limit, given E > 0 , 3 6, > 0 and 6, > 0 such that (f(x)-LI<&forOI~-aI<6,and Ih(x)-L)<eforO(x-aI<6,. 'Let 6 = min ( 4 , 6,). Then, O<(x-a1<6* If(x)-L(<&andIh(x)-LICE 3L-eIf(x)
Theorem 2 is also called the sandwich theorem (or the squeeze theorem), because g is being smdwiched between f and h. Let us see how this theorem can be used.
Example 4 Given that I f(x)- 1 ( 5 3(x +
t/ x E R, can we calculate
lim f (x) ?
x+-1
We know that-3(x + S f(x) -1 S 3 (x + v x. This means that -3(x + + 1 5 f(x) 5 3(x + 1)2+ 1 5 t~ x. Using the sandwich theorem and the fact that lirn [-3 (x+
x+-1
+ 1]= 1 = lirn [3(x+ I-'-.?
+ I], weget lirn f(x)= 1 x-'-
I
In the next section we will look at the limits of the sum, product and quotient of functions.
Algebra of limits
2.2.1
Now that you are familiar with limits. Let us state some basic properties of limits. (Their proofs are beyond and scope of this course.) 'Theorem3 Let f and g be two h c t i o n s such that lim f(x) and lim g(x) exist. Then P'-X
i)
P'-X
lim[f(x)+g(x)]=limf(x)+limg(x) P'-X
ii)
P'-X
lim [f(x) g (x)] = X-'P
5) lim X+P
[:yp
Sum rule
P 'x-
f(x)
]
lim g(x)
1 = I' ,provided lim g (x) gdx) litn g (x) x-rp
P'-X
Productrule t0
Reciprocal rule
We can easily prove two more rules in addition to the three rules given in Theorem 3. These are : *-' lirn k = k
iv)
Constant function rule
X+P
v)
limx=p
Identity function rule
x +P
We shall only indicate the method of proving iv) and v): iv) Here I f(x) - L I = 1 k - k I = 0 < E, whatever be the value of 8. *
Using the properties that we have just stated, we will calculate the limit in the following example. Example 5 Let us evaluate lim x+2
3x2 + 4x
2x + 1
NOW lim 2~ + 1 = lirn 2x + lim 1 by using i) x+2
x+2
lirn 2 lirn x + lirn 1 by using ii)
=
x+2
I
P
:.
x+2
x+2
= 2 x 2 + 1=5+Obyusingiv)andv) We can use (iii) of Theorem 3. Then the required limit is
lirn 3 x 2 + lirn 4x
lirn (3x2 + 4x)
x+2
lim ( 2 +~1)
!
x+2
x+2
-
x+2
lim 2 x + liml
x+2
x+2
b~u~ingi)
x+2
lirn 3 lim x lirn x + lim 4 lirn x
x+2
-
x+2
x+2
x+2
lim 2 lirn x + lim 1
x+2
x+2
x+2
by using ii)
x+2
3 x 2 x 2 + 4 x 2 = -20 - -4 2x2+1 5 You can similarly calculate the limits in the following exercises. -
E
E E3) t
3
E2) Show that lim - = 3 X+l
X
Calculate lirn x + l 2x +
(A)
Real Numbers and Functions
Elements o f Differential Calculus
2.2.2 Limits as x
-+
-
(or - -) 1
Take a look at the graph of the hnction f(x) = - ,x > 0 in Fig. 3. This is a decreasing function X
of x. In fact, we see fromFig. 3 that f(x) comes closer and closer to zero as x gets larger and larger. This situation is similar to the one where we have a hnction g(x) getting closer and closer to a value L as x comes nearer and nearer to some number p, that is when lim g(x) = L. X-'P
The only difference is that in the case of f(x), x is not approaching any finite value, and is just becoming larger and larger. We express this by saying that f(x) + 0 as x + w ,or lim f(x) = 0. X--W
Note that, = is not a real number. We write x + w merely to indicate that x becomes larger and larger. We now formalise this discussion in the following definition. Fig. 3
Definition 2 A function f is said to tend to a limit L as x tends to w if, for each E > 0 it is possible to choose K > 0 such that I f(x) - L I < E whenever x > K. In this case, as x gets larger and larger, f(x) gets nearer and nearer to L. We now give another example of this situation. Example 6 Let f be defined by setting f(x) = 1/x2for all x E R \ (0). Here f is defined for all real values of x other than zero. Let us substitute larger and larger values of x in f(x) = 1/x2and see what happens (see Table 2). Table 2
We see that as x becomes larger and larger, f (x) comes closer and closer to ie:o: Now, let us choose any E > 0. If x > l! J;, then 1/x2< E. Therefore, by choosing K = 11 &, we find that x >K
i f(x) / < E. Thus, lim
f(x) = 0.
X-SQ,
Fig. 4 gives us a graphic idea of how this function behaves as x + w.
AY
Fig. 4
Sometimes we also need to study the behaviour of a function f(x), as x takes smaller and smaller negative values. This can be examined by the following definition. Definition 3 A function f is said to tend to a limit L as x + - if, for each E > 0,it is possible to choose K > 0, such that / f(x) - L I < E whenever x < - K. The following example will help you in understanding this idea. Example 7 Consider the function f : R + R defined by
The graph o f f is as shown in Fig. 5.
LImltr and Continuity
What happens to f(x) as x takes smaller and smaller negative values? Let us make a table (Table 3) to get some idea. Table 3
We see that x takes smaller and smaller negative values, fix) comes closer and closer to zero. In fact 141 + x2)< E whenever 1 + x2> I/&,that is, whenever x2> (I/&)- 1, that is, whenever either
x < - k 3 I f(x) j C E. Consequently, lim fix) = 0. X+-@
In the above example we also find that lim f (x) = 0. x-bm
Let us see how lim f(x) can be interpreted geometrically.
-
x+m
In the above example, we have the function f(x) = 141 + x2),and as x -+ m, or x -+ - m, f(x) -+0. From Fig. 5 you can see that, as x -+ or x -+ - m ,the curve y = f(x) comes nearer and nearer the straight line y = 0, which is the x-axis. Similarly, if we say that lim g(x) = L, then it means that,as x -,m the curve y = g (x) comes x+m
closer and closer to the straight line y = L. x - 1. Example 8Let us show that lim -- X-'m (1 + x2)
I
Now, 1 x2/(1+ x2)- 1 1 = 141 + x2). In the previous example we have shown that 1 1/(1+ x2) ) < E for x > K, where K = 1 I/€- 1.' 1 Thus, given E > 0, we choose K = / 1 / -~1 JIJZ, so that
We show this geometrically inFig. 6. 4y
1
Fig. 6
-
Elements o f Differential Calculus
You must be wondering if all the properties given in Theorem 3 also hold when we take limits as x + . Yes, they do. You can use them to solve this exercise.
E
E4) Shos that
a)
lirn l/x=O x-bm
b)
lim (l/x+3/x2+5)=5 X-bco
Sometimes we cannot use Theorem 2 directly, as is clear in the following example, Letus see how to overcome this problem. 3x + 1 Example 9 Suppose, we want to find xlirn -, 2x + 5
'
-
We cannot apply Theorem 3 directly since the limits of the numerator and the denominator, as x + ,cannot be found. Instead, we rewrite the quotient by multiplying the numerator and denominator by llx. for x#O. 3x+l - 3+(l/x) Then, 2 x - 2 + (5 I x) , for x # 0. Now we use Theorem 3 and the fact that lirn llx = 0 (see # 4 a), to get x-bm
3 + (1 / x) 3x + 1 lim = lim x-+m2x+5 x - + m 2 + ( 5 / x )
-
lim ( 3 + l / x ) x-+m
- lirn ( 2 + 5 / x ) x-+m
E
3+0
- -=-
2+0
-
2
By now you knust be used to the various defmitions of limits, so can try this exercise E5) a) If for some E > 0,and for ever K, 3 x >K s.t. I f(x) - L 1 > E, what will you infer? b) If lirn f(x) ;f L, how can you express it in the E - 6 form? x -tP
Real Numbers and Functions
We end this section with the following important remark. Remark 2 In case we have to show that a function f does not tend to a limit L as x approaches p, we shall have to negate the definition of limit (also see E5(b)). Let us see what this means. Suppose we want to prove that lim f(x) ;t L. Then, we should find some E > 0 such that for every 6 > 0, there is some x E ]p Y8for which I (x) - L I > 6. Through our next example we shall illustrate the negation of the definition*ofthe limit of f(x) as x +00. Example 10 To show that lirn llx # I, we have to find some E 0 such that for any K (howsoever large) we can alGzys find an x > K such that ( I/x - 1 I > E. 114. This clearly show that lim l/x # 1. X+(D
2.2.3
One-sided Limits
If we consider the graph of the function f(x) = [XI,shown in Fig. 7, we see that f(x) does not see to approach any fixed value as x approaches 2. But from the graph we cansay that if x approaches 2 from the left then f(x) seems to tend to 1. At the same time, if x approaches 2 from the right, then f(x) seems to tend to 2. This means that the limit o f f cxists if x approaches 2 from only onc side (left or right ) at a time. This example suggests that we in!roduce the idea of a one-sided limit. Definition 4 Let f be a function defined for all x in the interval ]p, q[. f is said to approach a limit L as x approaches p from right if, given any E > 0, there exist a number 6 > 0 such that p<x
Similarly, the function f : )a, p[ + R is said to approach a limit L as x approaches p f r o p the left i f , g i v e n a n y ~ > 0 , 3 6~ O s u c h t h a t p - - 6 < x < p ~ I f ( x ) - L I < & . This limit is denoted by lirn f (x). x+p-
Note that in computing these limits the values of f(x) for x lying on only one side of p are taken ~ n t oaccount. Let uf apply this definition to the function f(x) = [ x 1, we know that for x E [l, 2[, [x] = 1. That is, [x] is a constant function on [I, 2[. Hence lim [x] = 1. Arguing similarly, we find that since [x] x+2-
=:
2 for all x E [2,3[, [XIis, again, aconstant function on [2,3[, and lim [x] = 2. x+2+
Let us improve our understanding of the definition of one-sided limits by lookiilg at some more examples. u
Example 11 Let f be defined on R by setting
We will show that lirn f(x) equals -1. x 4 -
When x < 0 , )x 1 = - x, and therefore, f(x) = (- x)lx = -1. In order to show that lim f(x) exists x-0-
and equals -1, we have to start with any E > 0 and then find a 6 > 0 such that, if
-6<x
Example 12 f is a function defined on R by setting f(x)= x - [x], for all x E R. Let us examine whether lim f(x) exists. x+1-
Recall (Unit 1) that this function is given by f(x) = x, if 0 5 x < 1. f(x) = x - 1 if 1 5 x < 2, and, in general
= -1.
Fig. 7
Elements of Differential Calculus
Fig. 8
Since f(x) = x for values of x less than 1 but closc to 1, it is reasonabie to expect that lim f(x) = 1. Let us prove this by taking any E : O and choosing 6 = min / 1, E I. We f h d x-1-
1-6<x
I=Ix-l ) < 6 < & .
Therefore. l i n ~ f(x) = 1. X-+l
Proceeding exactly as above, the noting that f(x) = x - 1 if 1 5 x < 2, we can similarly prove that f(x) =O.
E
5 6 ) Prove that
a)
lim x [x] -4 x-3-
b)
lim -1x1 -=I
n+Oi
X
Before going further, !et us see how the concepts of one-sided limit and limit are comected.
Real
Theorem 4 'The following statements are equivalent. lirn qx)
i)
exists
X+P
lirn f ( X )
ii)
and l i n ~ exist and are equal. x-+p-
Proof :'To show that i) and ii) are equivalent, we have to show that i) 3 ii) and ii) s i). We first prove that i) 3 ii). For this we assume that lirn f(x) = L. Then given *E > 0.
"36>@suchthat]f(x)-L)<~forO
X-+P
lirn f ( x ) = L z lim f(x). x+p-
x+p+
We now prove the converse, that is, ii) 3 i). For this, we assume that
lirn f ( x ) :- linl f(x) = L. Then, givcii E > 0,36, ,6, > 0. x+p-
x+pt
such that if(x)-Li<~forp-6,<x
Hence, lirn fjx) =L. X+P
Thus, we have shown that i) =, ii) and ii) 3 i), proving that they are equivalent. From Theorem 4, we can conclude that if lim qx) exists, then lirn f(x) and Aim f(x) x-' P x-tp* x-tpalso exist and further. lim f(x)= lirn f(x)= lirn f(x) x
P
x-tp+
-+
x4p-
Remark 3 If you apply Theorem 4 to the function f(x) = x - [x] (see Example 12), you will see that lim {x - [x]} does not exist as lirn (x - [x]) # lim {x- [x]} , x-1
x+1+
x+l-
We shall use this concept of one-sided limits to define continuous functions in the nest section.
2.3 CONTINUITY A continuous process is one that goes on smoothly without any abrupt change. Continuity of a function can also be interpreted in a similar way. Look at Fig. 9. The grapi? of the function f in Fig. 9 (a) has an abrupt cut at the point x = 3, whereas the graph of the function g in Fig. 9 (b) proceeds smoothly. We say that the function g is continuous, while f is not.
~4
y
4
(a)
(b) Fig. 9 : (a) Graph of f (b) Graph of g
Continuous functions play a very important role in calculus. As you proceed, you will be able to see that many theorems which we have stated in this course are true only for continuous functions. You will also see that continuity is a necessary condition for the derivability of a function, and that it is a sufficient condition for the integrability of a function. Bilt let us give a precise meaning to "a continuous function'' now.
umbers
and Functions
E i e n ~ e n t sof Differential Calculus
Definitions and Examples
2.3.1
In this sectLon we shall give you the definition and some examples of a continuous function. We shall also give you a short list of conditions which a function must satisfy in order to be continuous at a point.
Definition 5 Let f be a function defined on a domain D, and let r be a positive real number such that the interval ]p - r, p + r[ c D. f is said to be continuous at x = p if lim f(x) = f(p). X+P
,
By the definition of lim~tthls means that f is continuous at p is given E > 0 , 3 6 > 0 such that I fix)
- f(p) ) < F whenever 1 x - p j < 6. To clarify this concept let us look at an example
'
Example 13 L.et us check the continuity of the function
-
f : L R 4 W such that f(x) x at the point x = 0. Now, f(0) = 0. Thus we want to know if lirn f (x) = 0. x+O
This is true because given E > 0,,we can choose 6 = E and verify that ( x ( < 6 3 I f (x) I < E. Thus f is aontinuous at x = 0.
Remark 4 f is cont~nuousat x = p provided the following two criteria are met : i)
.
lim f(x)
x+p
exists.
Fig. 10 : (a) Graph of f
(b) G r a ~ hof g
Fig. 1U ~lioiwstwo discontinuous functions f and g. Criterion i) IS not inet by f, whereas g fails to meet criterion ii). If you read Remark 3 again, you will find that f(x) = x - [x] is not continuous as x = 1. But we have see that we can calculate one-sided limits of f(x) = x - [x] at x = 1. This leads us to the following definition.
Dennition 6 A function f : lp, q[ R is said to be continuous from the right at x == p ~f lirn f(k) = f (p). We say that f is continuous from the left at q if lirn f(x) = f (q). x+pt
x+q-
Thus, f(x) = x - [xl is continuous from the right but not from the left at x = 1 since lim f ( x ) x-1-
it f(1) and
lirn f(x)- f(1) =0. x+l'
E E 7) Give E - 6 definition of continuity at a point from the right as well as from the le'ft.
E
E 8) Show that function f: ]p - r, ,p + r[ -+ R is continuous at x = p if and only i f f is continuous from the right as well as from the left at x p (Use Theorem 4).
Real Numbers and Functions
;-
Now that you know how to test the continuity of a function at a point, let us go a step further, and define continuity of a function on a set.
Definition 7 A function f defined on a domain D is said to be continuous on D, if it is continuous at every point of D. Let us see some more examples. Example 14 Let f(x) = xn for all x E R and any n E Zt. Show that f(x) is continuous at x = p for all p E R. We know that lirn x = p for any p E R. Then, by the product rule in Theorem 3, we get
z' is the set of positive integers
X+F
lirn xn = ( lirn x) ( lirn x) ..........( lim x) X+P
-
x+p
x+p
(n times)
X+P
pp ............. p (n times) =p". Therefore, lirn f(x) exists and equals f(p). Hence f is x+p. continuous at x = p. Since p was any arbitrary number in R, we can say that f is continuous on R. Remark 5 Using Example 14 and Theorem 3, we can also prove that polynomial a, + alx + ............ + allx", where a,, a, ..............an E R, is continuous on R, that is, l i m ( a o + a l x + ................. + a,,xn)= a, + alp + ..............+ anpnfor all p E R. -
X-tP
Example 15 The greatest integer function f : R -+ R : f(x) = [ x ] is cliscontinuous at x = 2. To prove this we recall our discussion in Sec. 2 in which we have proved that lim f(x) = 1 x+2-
and lirn f(x) = 2. Thus, since these two limits are not equal lirn f(x) does not exist. x+2+
-
x+2
Therefore, f is not continuous at x = 2 because the first criterion laid down in Remark 4 is not met. Example 16 Let f(x) = / x / for all x E R. This f is continuous at x = 0. Here f(x) = x, if x 2 0, and f(x) = - x if x < 0. You can show that lirn f(x)= lim x=O=f(O)and
C
x+o*
x+O'
lim f(x) = lirn (-x)
x+o
f
1
x+o-
= 0 = f(0). Thus
lirn f(x) exists and equals f(0). Hence f is continuous
x+o
atx=O. Note : f is also continuous at every other point of R. (Check this statement). x2 - 1
Example 17 Suppose we want to find whether f(x) = -is continuous at x = 0. x-1 In Fig. 11 you see the graph o f f . It is the line y = x + 1 except for the point (1,2) Y'
Fig. 11
We can write f(x) all x 1.'
=
x
+ 1 for
Elements of Dlfferentlal
f(o)=(02- 1)/(0- I ) = 1 and limf(x)= lim (x2-l)/(x-1)=
Calculus
x-ro
1
x-to
lim ( x i l ) = l
x+o
lim f(x) = f(O), so that f is continuous at x = 0.
x+O
The exponential function f(x) =ex and the logarithmic h c t i o n f(x) = In x are continuous functions. You can check this by looking at their graphs in Unit 1. Similarly,x +sinx and x + cosx are continuous, x + tanx is continuous in ] - d 2 , n;2[. This fact is quite obvious from the graphs of these functions (We have given their graphs in Unit 1). We shall not attempt a rigorous proof of their continuity here. Caution :Checking the continuity of a function from the smoothness of its graph is not a fool-proof method. If you look at the graph (Fig. 12) ofthe h c t i o n x +x sin (llx), you will find that it has no breaks in the neighbourhood of x = 0. But this function is not continuous. Observe that the graph oscillates wildly near zero.
t
. Fig.
E
12
E9) Show that the function f : R + R given by f(x) = 1!(x2 - 9) is continuous at all points of Rexceptatx=3andx=-3.
Now that we know how to check whether a function is continuous or not, let us go further, and talk about the continuity of some combinations of functions.
2.3.2 Algebra of Continuous Functions Let f and g be functions defined and continuous on a common domain D E R, and let k be any real number. In Unit 1, we defied the functions f + g, fgl Ug (provided g(x) z 0 any where in D), kif and I f I. The following theorem tells us about the continuity of these functions.
Theorem 5 Let f and g be functions defined and continuous on a common domain D, and let k be any real number. The functions f + g, kf, 1 f I and fg are all continuous on D. If g(x) # 0 anywhere in D, then the function flg is also continuous on D. We shall not prove this theorem here. In Unit 1, you have studied the important concept of composite functions. In Theorem 6, we will talk about the continuity of the composite of two continuous functions. Here again, we shall state the theorem without giving proof as it is beyond the level of this course.
Theorem 6 Let f : D, + D2and g :D2 + D, be continuous on their domains. Then gof is continuous on Dl, (Dl, D2,D3c R). Example 18 To prove that f : R +R : f (x) = (x2+ is continuous at x = 0, we consider the functions g : R +R : g(x) = x3 and h : R +R :h(x) = x2+ 1. You can check that f(x) = goh (x). Further, by Remark 5, h is continuous on R, and g is also continuous on R. Thus, goh = f is continuous on R. Let us see if the converse of the above theorems are true. For example, iff and g are defined on an interval [a, b] and iff + g is continuous on [a, b], does that mean that f and g are continuous on [a, b] ? No. Consider the functions f and g over the interval [0, 11 given by
Then neither f nor g is continuous at x = 112. (Why?) But (f + g) (x) = 1 v x E [O, 11. Therefore, f + g is continuous on [0, 11. Now, if ( f I is continuous at a point p, must f also be continuous at p? Again, the answer is No.Take, for example, the function f : R +R givenby f(x) =
-1, 1,
forx1O for > 0
Then I f(x) 1 = 1 in Rand hence I f 1 is continuous. But f is not continuous at x = 0 (Why?) 1,
E10) Iff : R +R is defined by f (x) = f(x) = is f continuous at a ) x = 1
C
b) x = -3/2?
-1.
i f x ~ Z ifxeZ
Limits and Continuity
,
E l e m e ~ ~ tosf Differential Calculus
One again, we won't prove this theorem here. But try to understand its statement because we shall be using it in subsequent units. Theorem 7 (Intermediate Value Theorem) Let f be continuous on the closed interval [a, b]. Suppose c ia a real number lying between f(a) and f(b). (That is, f(a) < c < f(b) or f(a) > c > (b?). Then there exists some x, E ]a, b[, such that f(xo)= c. How can we interpret this geometrically? We have already seen that the graph of a continuous function is smooth. It does not have any breaks or jumps. This theorem says that, if the points (a, f(a)) and (b, f(b)) lie on two opposite sides of a lide y = c (see Fig. 13), then the graph o f f must cross the line y = c.
Fig. 13
Note that this theorem guarantees only the existence of the number x,. It does not tell us how to find it. Another thing to note is that this x, need not be unique. That brings us to the end of this unit.
SUMMARY
2.4
Wh end this untt by summarising what we have covered in it.
.
The limit of a function f a t a point p of its domain is L is given E > 0 , 3 6 > 0, such that )f(x)-L I <&whenever1x-pI<6. One-sided limits
3.
p,,
1.
lim f(x) exists if and only if lim
x+p+
4.
f (x)and? -,, lim
f(x) both exist and are equal.
A fuqction f is continuous at a point x = p if lim f (x) = f (P)
,,
5.
6.
If the function f and g are continuous on D, then so are the functions f + g, fg, ( f 1, kf (where k E R) flg (where g(x) # 0 in D) The Intermediate Value Theorem : Iff is continuous on [a, b] and if f(a) < c < f(b) (or f(a) > c > f(b)), then 3x0€ ]a, b[suchthat f(xo)=c. 4
2.5
QOLUTIONS AND ANSWERS
E 1) a) Given any E > 0,if we choose 6 = min {d2,1/2) then I ~ - l I < 6 < 1 / 2 - x>1/2.
Real Numbers and Functions
Hence lim l/x = 1 x+l
Given E > 0, ifwe choose 6
= min
1 (217) E, 1/21,then
Ix-I(<1/2*~<3/2*~+2<7/2and
x3 - 1 Hence, lim -= 3 x+l x - 1 lim 3 1 E2) lirn 3 / x = + = 3 / 1 = 3 x+l hm x x+l
E3) lirn 2 x + 5 x+l
f\ I + x ) =21im x + 15+lirnl i mx2x 2 x2
X+I
E4) a) Given e > 0 if we choose K = lle, then x > K - I l/x-OI=) l / x I < l / K = ~ Thus, lirn l/x=O x+m
b) Given E > 0, if we choose K = 11& ,then x > K * 1 1/x2-OI=( l / x 2 ( i : 1 / K 2 = ~ . Hence, lirn 1/x2=O x+m
Now, lirn (l/x +.3/x2+ 5) x+m
=
lirn 1/x+3 lirn 1/x2+ lirn 5 = 0 + 3 x 0 + 5 = 5 x+m
x+m
E5) a)
x+m
lirn f(x)#L x+m
lirn X-[XI=
x+3-
b)
lirn X+O+
;;y-
x-2=l
I X \ / X J= ~:J+' X / X = ~ , I X I = X ~ O ~ X > O .
- x+olim -(x2+2)=-2 E 7) f is continuous from the right at x = p if v E > 0 there exists a 6 > 0 s.t. p<~
* !!4x) = f(p)
lim f(x) = f (p) and x+plim * ,+,+
f (x) = f(p) by Theorem 4.
f is continuous from right and from left at x = p. Iff is continuous from right and left,
Elements of Differential Calculus .
lim 3
!$ fix) exists and = qp)
3f
is continuous at p,
E9) lim f(x) = x-+a
lim x2 - 9
--
1
a2 - 9
= f (a) for all a except a = 3 f 3
X+l
Hence f is continuous at all points except at f 3, f is not defined at 23. E10) a) f is not continuous at x = 1. For E = 1 and any 8 > 0, if x is any non-integer E ] 1-8,1+8[,then ( f l ~ ) - f ( l ) ( = I - l - lI = ~ > E . b) f is continuous at -312. Since given E > 0, if we choose 8 < 112, then
Ix-(-312)1< 1 1 2 3 - 2 < x < - 1 3 x 4 Zandhence )f(x)-f(-312) /=O<E.
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