MAL 100: Calculus

Lecture Notes

2 Continuity, Dierentiability and Taylor's theorem

2.1 Limits of real valued functions

Let f(x) be dened on (a; b) except possibly at x0.

Denition 2.1.1. We say that limx!x0

f(x) = L if, for every real number > 0; there exists

a real number > 0 such that

0 < jx x0j < =) jf(x) Lj < : (2.1)

Equivalently,

Remark 2.1. The above denition is equivalent to: for any sequence fxng with xn ! x0,we have f(xn)! L as n!1.Proof. Suppose limx!x0f(x) exists. Take > 0 and let fxng be a sequence convergingto x0. Then there exists N such that jxn x0j < for n N . Then by the denitionjf(xn) Lj < . i.e., f(xn)! L.For the other side, assume that xn ! c =) f(xn) ! L. Suppose the limit does notexist. i.e., 9 0 > 0 such that for any > 0; and all jx x0j < , we have jf(x)Lj 0.Then take = 1

nand pick xn in jxn x0j < 1n , then xn ! x0 but jf(xn) Lj 0. Not

possible.

Theorem 2.1.2. If limit exists, then it is unique.

Proof. Proof is easy.

Examples: (i) limx!1

(3x

2 1) = 1

2. Let > 0. We have to nd > 0 such that (2.1) holds

with L = 1=2. Working backwards,

3

2jx 1j < whenever jx 1j < := 2

3:

(ii) Prove that limx!2

f(x) = 4, where f(x) =

(x2 x 6= 21 x = 2

Problem: Show that limx!0

sin(1

x) does not exist.

Consider the sequences fxng = f 1ng; fyng = f 1

2n + 2

g. Then it is easy to see that

1

xn; yn ! 0 and sin

1

xn

! 0; sin

1

yn

! 1. In fact, for every c 2 [1; 1]; we can nd

a sequence zn such that zn ! 0 and sin( 1zn)! c as n!1:

By now we are familiar with limits and one can expect the following:

Theorem 2.1.3. Suppose limx!c

f(x) = L and limx!c

g(x) = M , then

1. limx!c

(f(x) g(x)) = LM .

2. f(x) g(x) for all x in an open interval containing c. Then L M .

3. limx!c

(fg)(x) = LM and when M 6= 0, limx!c

f

g(x) =

L

M:

4. (Sandwich): Suppose that h(x) satises f(x) h(x) g(x) in an interval contain-ing c, and L = M . Then lim

x!ch(x) = L.

Proof. We give the proof of (iii). Proof of other assertions are easy to prove. Let > 0.

From the denition of limit, we have 1; 2; 3 > 0 such that

jx cj < 1 =) jf(x) Lj < 12

=) jf(x)j < N for some N > 0;

jx cj < 2 =) jf(x) Lj < 2M

; and

jx cj < 3 =) jg(x)M j < 2N

:

Hence for jx cj < = minf1; 2; 3g, we have

jf(x)g(x) LM j jf(x)g(x) f(x)M j+ jf(x)M LM j jf(x)jjg(x)M j+M jf(x) Lj< :

To prove the second part, note that there exists an interval (c ; c+ ) around c suchthat g(x) 6= 0 in (c ; c+ ).Examples: (i) lim

x!0xm = 0 (m > 0): (ii) lim

x!0x sinx = 0:

Remark: Suppose f(x) is bounded in an interval containing c and limx!c

g(x) = 0. Then

limx!c

f(x)g(x) = 0:

2

Examples: (i) limx!0

jxj sin 1x= 0. (ii) limx!0 jxj ln jxj = 0:

One sided limits: Let f(x) is dened on (c; b): The right hand limit of f(x) at c is L,

if given > 0, there exists > 0, such that

x c < =) jf(x) Lj < :

Notation: limx!c+

f(x) = L. Similarly, one can dene the left hand limit of f(x) at b and is

denoted by limx!b

f(x) = L:

Both theorems above holds for right and left limits. Proof is easy.

Problem: Show that lim!0 sin = 1.Solution: Consider the unit circle centred at O(0; 0) and passing through A(1; 0) and

B(0; 1). Let OT be the ray with \AOT = ; 0 < < =2. Let P be the point ofintersection of OT and circle. Then OPQ and OTA are similar triangles and hence,

Area of OAP < Area of sector OAP < area of OAT . i.e.,

1

2sin cos . Now lim

!0+cos = 1 implies that lim

!0+sin

= 1.

Now use the fact thatsin

is even function.

B

OQ A

P

At this stage, it is not dicult to prove the following:

Theorem 2.1.4. limx!a f(x) = L exists () limx!a+ f(x) = limx!a f(x) = L.

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Limits at innity and innite limits

Denition 2.1.5. f(x) has limit L as x approaches +1, if for any given > 0, thereexists M > 0 such that

x > M =) jf(x) Lj < :Similarly, one can dene limit as x approaches 1.Problem: (i) lim

x!11

x= 0, (ii) limx!1 1x = 0: (iii) limx!1

sinx does not exist.

Solution: (i) and (ii) are easy. For (iii), Choose xn = n and yn =2+ 2n. Then

xn; yn !1 and sin xn = 0, sin yn = 1. Hence the limit does not exist.

Above two theorems on limits hold in this also.

Denition 2.1.6. (Horizontal Asymptote:) A line y = b is a horizontal asymptote of

y = f(x) if either limx!1

f(x) = b or limx!1

f(x) = b.

Examples: (i) y = 1 is a horizontal asymptote for 1 + 1x+1

Denition 2.1.7. (Innite Limit): A function f(x) approaches 1 (f(x)!1) as x!x0 if, for every real B > 0, there exists > 0 such that

0 < jx x0j < =) f(x) > B:

Similarly, one can dene for 1. Also one can dene one sided limit of f(x) approaching1 or 1.Examples (i) lim

x!01

x2=1, (ii) limx!0 1x2 sin( 1x) does not exist.

For (i) given B > 0, we can choose 1pB: For (ii), choose a sequence fxng such

that sin 1xn

= 1, say 1xn

= 2+ 2n and 1

yn= n: Then lim

n!1f(xn) =

1

x2n! 1 and

limn!1

f(yn) = 0; though xn; yn ! 0 as n!1.

Denition 2.1.8. (Vertical Asymptote:) A line x = a is a vertical asymptote of y = f(x)

if either limx!a+

f(x) = 1 or limx!a

f(x) = 1:

Example: f(x) = x+3x+2

.

x = 2 is a vertical asymptote and y = 1 is a horizontal asymptote.

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2.2 Continuous functions

Denition 2.2.1. A real valued function f(x) is said to be continuous at x = c if

(i) c 2 domain(f)(ii) lim

x!cf(x) exists

(iii) The limit in (ii) is equal to f(c).

In other words, for every sequence xn ! c, we must have f(xn) ! f(c) as n ! 1.i.e., for a given > 0, there exists > 0 such that

0 < jx cj < =) jf(x) f(c)j < :

Examples: (i) f(x) =

(x2 sin( 1

x) x 6= 0

0 x = 0is continuous at 0.

Let > 0. Then jf(x) f(0)j jx2j. So it is enough to choose = p.

(ii) g(x) =

(1xsin 1

xx 6= 0

0 x = 0is not continuous at 0.

Choose 1xn

= 2+ 2n. Then lim xn = 0 and f(xn) =

1xn!1:

The following theorem is an easy consequence of the denition.

Theorem 2.2.2. Suppose f and g are continuous at c. Then

(i) f g is also continuous at c(ii) fg is continuous at c

(iii)f

gis continuous at c if g(c) 6= 0.

Theorem 2.2.3. Composition of continuous functions is also continuous i.e., if f is

continuous at c and g is continuous at f(c) then g(f(x)) is continuous at c.

Corollary: If f(x) is continuous at c, then jf j is also continuous at c.Theorem 2.2.4. If f; g are continuous at c, then max(f; g) is continuous at c.

Proof. Proof follows from the relation

max(f; g) =1

2(f + g) +

1

2jf gj

and the above theorems.

Types of discontinuities

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Removable discontinuity: f(x) is dened every where in an interval containing a ex-

cept at x = a and limit exists at x = a OR f(x) is dened also at x = a and limit is NOT

equal to function value at x = a. Then we say that f(x) has removable discontinuity at

x = a. These functions can be extended as continuous functions by dening the value of

f to be the limit value at x = a.

Example: f(x) =

(sinxx

x 6= 00 x = 0

: Here limit as x! 0 is 1. But f(0) is dened to be 0:

Jump discontinuity: The left and right limits of f(x) exists but not equal. This type

of discontinuities are also called discontinuities of rst kind.

Example: f(x) =

(1 x 01 x 0. Easy to see that left and right limits at 0 are dierent.

Innite discontinuity: Left or right limit of f(x) is 1 or 1.

Example: f(x) = 1xhas innite discontinuity at x = 0.

Discontinuity of second kind: If either limx!c

f(x) or limx!c+

f(x) does not exist, then c

is called discontinuity of second kind.

Properties of continuous functions

Denition 2.2.5. (Closed set): A subset A of IR is called closed set if A contains all its

limit points. (i:e:, if fxng A and xn ! c, then c 2 A).Theorem 2.2.6. Continuous functions on closed, bounded interval is bounded.

Proof. Let f(x) be continuous on [a; b] and let fxng [a; b] be a sequence such thatf(xn) ! 1. Then fxng is a bounded sequence and hence there exists a subsequencefxnkg which converges to c. Then f(xnk)! f(c); a contradiction.

Theorem 2.2.7. Let f(x) be a continuous function on closed, bounded interval [a; b].

Then maximum and minimum of functions are achieved in [a; b].

Proof. Let fxng be a sequence such that f(xn)! max f . Then fxng is bounded and henceby Bolzano-Weierstrass theorem, there exists a subsequence xnk such that xnk ! x0 forsome x0. a xn b implies x0 2 [a; b]. Since f is continuous, f(xnk) ! f(x0). Hencef(x0) = max f . The attainment of minimum can be proved by noting that f is alsocontinuous and min f = max(f).

6

Remark: Closed and boundedness of the interval is important in the above theorem.

Consider the examples (i) f(x) = 1xon (0; 1) (ii) f(x) = x on IR.

Theorem 2.2.8. Let f(x) be a continuous function on [a; b] and let f(c) > 0 for some

c 2 (a; b), Then there exists > 0 such that f(x) > 0 in (c ; c+ ).Proof. Let = 1

2f(c) > 0. Since f(x) is continuous at c, there exists > 0 such that

jx cj < =) jf(x) f(c)j < 12f(c)

i.e., 12f(c) < f(x) f(c) < 1

2f(c). Hence f(x) > 1

2f(c) for all x 2 (c ; c+ ).

Corollary: Suppose a continuous functions f(x) satisesR baf(x)(x)dx = 0 for all con-

tinuous functions (x) on [a; b]. Then f(x) 0 on [a; b].Proof. Suppose f(c) > 0. Then by above theorem f(x) > 0 in (c ; c + ). Choose(x) so that (x) > 0 in (c =2; c + =2) and is 0 otherwise. Then R b

af(x)(x) > 0. A

contradiction.

Alternatively, one can choose (x) = f(x).

Theorem 2.2.9. Let f(x) be a continuous function on IR and let f(a)f(b) < 0 for some

a; b. Then there exits c 2 (a; b) such that f(c) = 0.Proof. Assume that f(a) < 0 < f(b). Let S = fx 2 [a; b] : f(x) < 0g. Then [a; a+ ) Sfor some > 0 and S is bounded. Let c = supS. We claim that f(c) = 0. Take

xn = c +1n, then xn 62 S, xn ! c. Therefore, f(c) = lim f(xn) 0. On the otherhand,

taking yn = c 1n , we see that yn 2 S for n large and yn ! c, f(c) = lim f(yn) 0.Hence f(c) = 0.

Corollary: Intermediate value theorem: Let f(x) be a continuous function on [a; b]

and let f(a) < y < f(b). Then there exists c 2 (a; b) such that f(c) = y

Remark: A continuous function assumes all values between its maximum and minimum.

Problem: (xed point): Let f(x) be a continuous function from [0; 1] into [0; 1]. Then

show that there is a point c 2 [0; 1] such that f(c) = c:

Dene the function g(x) = f(x) x. Then g(0) 0 and g(1) 0. Now Apply Interme-diate value theorem.

7

Application: Root nding: To nd the solutions of f(x) = 0, one can think of dening

a new function g such that g(x) has a xed point, which in turn satises f(x) = 0.

Example: (1) f(x) = x3 + 4x2 10 in the interval [1; 2]. Dene g(x) = 104+x

1=2. We can

check that g maps [1; 2] into [1; 2]. So g has xed point in [1; 2] which is also solution

of f(x) = 0. Such xed points can be obtained as limit of the sequence fxng, wherexn+1 = g(xn); x0 2 (1; 2). Note that

g0(x) =

p10

(4 + x)3=2 0, there exists > 0 such that

x; y 2 S; jx yj < =) jf(x) f(y)j < :

Here depends only on , not on x or y.

Proposition: If f(x) is uniformly continuous function () for ANY two sequencesfxng; fyng such that jxn ynj ! 0, we have jf(xn) f(yn)j ! 0 as n!1.

Proof. Suppose not. Then there exists fxng; fyng such that jxn ynj ! 0 and jf(xn) f(yn)j > for some > 0. Then it is clear that for = , there is no for whichjx yj < =) jf(x) f(y)j < . Because the above sequence satises jx yj < , butits image does not.

For converse, assume that for any two sequences fxng; fyng such that jxnynj ! 0 wehave jf(xn) f(yn)j ! 0. Suppose f is not uniformly continuous. Then by the denitionthere exists 0 such that for any > 0, jx yj < =) jf(x) f(y)j > 0. Nowtake = 1

nand choose xn; yn such that jxn ynj < 1=n. Then jf(xn) f(yn)j > 0. A

8

contradiction.

Examples: (i) f(x) = x2 is uniformly continuous on bounded interval [a; b].

Note that jx2 y2j jx+ yjjx yj 2bjx yj. So one can choose < 2b.

(ii) f(x) = 1xis not uniformly continuous on (0; 1).

Take xn =1

n+1; yn =

1n, then for n large jxn ynj ! 0 but jf(xn) f(yn)j = 1.

(iii) f(x) = x2 is not uniformly continuous on IR.

Take xn = n+1nand yn = n. Then jxn ynj = 1n ! 0, but jf(xn) f(yn)j = 2+ 1n2 > 2:

Remarks:

1. It is easy to see from the denition that if f; g are uniformly continuous, then f gis also uniformly continuous.

2. If f; g are uniformly continuous, then fg need not be uniformly continuous. This

can be seen by noting that f(x) = x is uniformly continuous on IR but x2 is not

uniformly continuous on IR.

Theorem 2.2.10. A continuous function f(x) on a closed, bounded interval [a; b] is

uniformly continuous.

Proof. Suppose not. Then there exists > 0 and sequences fxng and fyng in [a; b] suchthat jxn ynj < 1n and jf(xn) f(yn)j > . But then by Bolzano-Weierstrass theorem,there exists a subsequence fxnkg of fxng that converges to x0. Also ynk ! x0. Now sincef is continuous, we have f(x0) = lim f(xnk) = lim f(ynk). Hence jf(xnk) f(ynk)j ! 0,a contradiction.

Corollary: Suppose f(x) has only removable discontinuities in [a; b]. Then ~f , the exten-

sion of f , is uniformly continuous.

Example: f(x) = sinxx

for x 6= 0 and 0 for x = 0 on [0; 1].Theorem 2.2.11. Let f be a uniformly continuous function and let fxng be a cauchysequence. Then ff(xn)g is also a Cauchy sequence.Proof. Let > 0. As f is uniformly continuous, there exists > 0 such that

jx yj < =) jf(x) f(y)j < :

Since fxng is a Cauchy sequence, there exists N such that

m;n > N =) jxn xmj < :

9

Therefore jf(xn) f(xm)j < .

Example: f(x) = 1x2

is not uniformly continuous on (0; 1).

The sequence xn =1nis cauchy but f(xn) = n

2 is not. Hence f cannot be uniformly

continuous.

2.3 Dierentiability

Denition 2.3.1. A real valued function f(x) is said to be dierentiable at x0 if

limh!0

f(x0 + h) f(x0)h

exists:

This limit is called the derivative of f at x0, denoted by f0(x0).

Example: f(x) = x2

f 0(x) = limh!0

2xh+ h2

h= 2x:

Theorem 2.3.2. If f(x) is dierentiable at a, then it is continuous at a.

Proof. For x 6= a, we may write,

f(x) = (x a)f(x) f(a)(x a) + f(a):

Now taking the limit x! a and noting that lim(x a) = 0 and lim f(x)f(a)(xa) = f

0(a), weget the result.

Theorem 2.3.3. Let f; g be dierentiable at c 2 (a; b). Then f g; fg and fg(g(c) 6= 0)

is also dierentiable at c

Proof. We give the proof for product formula: First note that

(fg)(x) (fg)(c)x c = f(x)

g(x) g(c)x c + g(c)

f(x) f(c)x c :

Now taking the limit x! c, we get the product formula

(fg)0(c) = f(c)g0(c) + f 0(c)g(c):

10

Since g(c) 6= 0 and g is continuous, we get g(x) 6= 0 in a small interval around c.Therefore

f

g(x) f

g(c) =

g(c)f(x) g(c)f(c) + g(c)f(c) g(x)f(c)g(x)g(c)

Hence

(f=g)(x) (f=g)(c)x c =

g(c)

f(x) f(c)x c f(c)

g(x) g(c)x c

1

g(x)g(c)

Now taking the limit x! c, we get

(f

g)0(c) =

g(c)f 0(c) f(c)g0(c)g2(c)

:

Theorem 2.3.4. (Chain Rule): Suppose f(x) is dierentiable at c and g is dierentiable

at f(c), then h(x) := g(f(x)) is dierentiable at c and

h0(c) = g0(f(c))f 0(c)

Proof. Dene the function h as

h(y) =

(g(y)g(f(c))

yf(c) y 6= f(c)g0(f(c)) y = f(c)

Then the function h is continuous at y = f(c) and g(y) g(f(c)) = h(y)(y f(c)), so

g(f(x)) g(f(c))x c = h(f(x))

f(x) f(c)x c :

Now taking limit x! c, we get the required formula.

Local extremum: A point x = c is called local maximum of f(x), if there exists > 0

such that

0 < jx cj < =) f(c) f(x):Similarly, one can dene local minimum: x = b is a local minimum of f(x) if there exists

> 0 such that

0 < jx bj < =) f(b) f(x):Theorem 2.3.5. Let f(x) be a dierentiable function on (a; b) and let c 2 (a; b) is a localmaximum of f . Then f 0(c) = 0.

11

Proof. Let be as in the above denition. Then

x 2 (c; c+ ) =) f(x) f(c)x c 0

x 2 (c ; c) =) f(x) f(c)x c 0:

Now taking the limit x! c, we get f 0(c) = 0:Theorem 2.3.6. Rolle's Theorem: Let f(x) be a continuous function on [a; b] and dier-

entiable on (a; b) such that f(a) = f(b). Then there exists c 2 (a; b) such that f 0(c) = 0.Proof. If f(x) is constant, then it is trivial. Suppose f(x0) > f(a) for some x0 2 (a; b),then f attains maximum at some c 2 (a; b): Other possibilities can be worked out similarly.Theorem 2.3.7. Mean-Value Theorem (MVT): Let f be a continuous function on [a; b]

and dierentiable on (a; b). Then there exists c 2 (a; b) such that

f(b) f(a) = f 0(c)(b a):

Proof. Let l(x) be a straight line joining (a; f(a)) and (b; f(b)). Consider the function

g(x) = f(x) l(x). Then g(a) = g(b) = 0. Hence by Rolle's theorem

0 = g0(c) = f 0(c) f(b) f(a)b a

Corollary: If f is a dierentiable function on (a; b) and f 0 = 0, then f is constant.

Proof. By mean value theorem f(x) f(y) = 0 for all x; y 2 (a; b).

Example: Show that j cos x cos yj jx yj.Use Mean-Value theorem and the fact that j sinxj 1.

Problem: If f(x) is dierentiable and sup jf 0(x)j < C for some C. Then, f is uniformlycontinuous.

Apply mean value theorem to get jf(x) f(y)j Cjx yj for all x; y.

Denition: A function f(x) is strictly increasing on an interval I, if for x; y 2 I withx < y we have f(x) < f(y). We say f is strictly decreasing if x < y in I implies

f(x) > f(y).

Theorem 2.3.8. A dierentiable function f is (i) strictly increasing in (a; b) if f 0(x) > 0for all x 2 (a; b). (ii) strictly decreasing in (a; b) if f 0(x) < 0.

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Proof. Choose x; y in (a; b) such x < y. Then by MVT, for some c 2 (x; y)

f(x) f(y)x y = f

0(c) > 0:

Hence f(x) < f(y).

2.4 Taylor's theorem and Taylor Series

Let f be a k times dierentiable function on an interval I of IR. We want to approximate

this function by a polynomial Pn(x) such that Pn(a) = f(a) at a point a. Moreover, if

the derivatives of f and Pn also equal at a then we see that this approximation becomes

more accurate in a neighbourhood of a. So the best coecients of the polynomial can

be calculated using the relation f (k)(a) = P(k)n (a); k = 0; 1; 2; :::; n. The best is in the

sense that if f(x) itself is a polynomial of degree less than or equal to n, then both f

and Pn are equal. This implies that the polynomial isnX

k=0

f (k)(a)

k!(x a)k. Then we write

f(x) = Pn(x) +Rn(x) in a neighbourhood of a. From this, we also expect the Rn(x)! 0as x! a. In fact, we have the following theorem known as Taylor's theorem:

pi pi/2 0 pi/2 pi4

3

2

1

0

1

2

3

4

pi x pi

sin

(x)

Plot of sin(x) and its Taylor polynomials

sin(x)

x

xx3/6

xx3/3!+x5/5!

Figure 1: Approximation of sin(x) by Taylor's polynomials

Theorem 2.4.1. Let f(x) and its derivatives of order m are continuous and f (m+1)(x)

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exists in a neighbourhood of x = a. Then there exists c 2 (a; x)( or c 2 (x; a)) such that

f(x) = f(a) + f 0(a)(x a) + :::::+ f (m)(a)(x a)m

m!+Rm(x)

where Rm(x) =f (m+1)(c)

(m+ 1)!(x c)m+1:

Proof. Dene the functions F and g as

F (y) = f(x) f(y) f 0(y)(x y) :::: f(m)(y)

m!(x y)m;

g(y) = F (y)x yx a

m+1F (a):

Then it is easy to check that g(a) = 0. Also g(x) = F (x) = f(x) f(x) = 0. Therefore,by Rolle's theorem, there exists some c 2 (a; x) such that

g0(c) = 0 = F 0(c) +(m+ 1)(x c)m

(x a)m+1 F (a):

On the other hand, from the denition of F ,

F 0(c) = f(m+1)(c)

m!(x c)m:

Hence F (a) =(x a)m+1(m+ 1)!

f (m+1)(c) and the result follows.

Examples:

(i) ex = 1 + x+x2

2!+x3

3!+ ::::+

xn

n!ec; c 2 (0; x) or (x; 0) depending on the sign of x.

(ii) sin x = x x3

3!+x5

5!+ ::::+

xn

n!sin(c+

n

2); c 2 (0; x) or (x; 0):

(iii) cos x = 1 x2

2!+x4

4!+ ::::+

xn

n!cos(c+

n

2); c 2 (0; x) or (x; 0).

Problem: Find the order n of Taylor Polynomial Pn, about x = 0 to approximate ex in

(1; 1) so that the error is not more than 0:005

Solution: We know that pn(x) = 1 + x+ ::::+xn

n!. The maximum error in [-1,1] is

jRn(x)j 1(n+ 1)!

max[1;1]

jxjn+1ex e(n+ 1)!

:

14

So n is such that e(n+1)!

0:005 or n 5.

Problem: Find the interval of validity when we approximate cosx with 2nd order poly-

nomial with error tolerance 104.

Solution: Taylor polynomial of degree 2 for cosx is 1 x22. So the remainder is (sin c)x

3

3!.

Since j sin cj 1, the error will be atmost 104 if jx33!j 104. Solving this gives jxj < 0:084

Taylor's Series

Suppose f is innitely dierentiable at a and if the remainder term in the Taylor's formula,

Rn(x)! 0 as n!1. Then we write

f(x) =1Xn=0

f (n)(a)

n!(x a)n:

This series is called Taylor series of f(x) about the point a.

Suppose there exists C = C(x) > 0, independent of n, such that jf (n)(x)j C(x). ThenjRn(x)j ! 0 if lim

n!1jx ajn+1(n+ 1)!

= 0. For any xed x and a, we can always nd N such

that jx aj < N . Let q := jxajN

< 1. Then(x a)n+1(n+ 1)! = jx aj1

jx aj2 ::: jx ajN 1

jx ajN ::: jx ajn+ 1

0.

Also the above observations show that if f 0(a) = 0; f 00(a) = 0 and f (3)(a) 6= 0, then thesign of f(x) f(a) depends on (x a)3. i.e., it has no constant sign in any intervalcontaining a. Such point is called point of inection or saddle point.

We can also derive that if f 0(a) = f 00(a) = f (3)(a) = 0, then we again have x = a is alocal minima if f (4)(a) > 0 and is a local maxima if f (4)(a) < 0.

Summarizing the above, we have:

Theorem 2.4.3. Let f be a real valued function that is dierentiable 2n times and f (2n)

is continuous at x = a . Then

1. If f (k)(a) = 0 for k = 1; 2; ::::2n 1 and f (2n)(a) > 0 then a is a point of localminimum of f(x)

2. If f (k)(a) = 0 for k = 1; 2; ::::2n1 and f (2n)(a) < 0 then a is a point local maximumof f(x).

3. If f (k) = 0 for k = 1; 2; ::::2n 2 and f (2n1)(a) 6= 0, then a is point of inection.i.e., f has neither local maxima nor local minima at x = a.

L'Hospitals Rule:

Suppose f(x) and g(x) are dierentiable n times, f (n); g(n) are continuous at a and

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f (k)(a) = g(k)(a) = 0 for k = 0; 1; 2; ::::; n 1. Also if g(n)(a) 6= 0. Then by Taylor'stheorem,

limx!a

f(x)

g(x)= lim

x!af (n)(c)

g(n)(c)

=f (n)(a)

g(n)(a)

In the above, we used the fact that g(n)(x) 6= 0 "near x = a" and g(n)(c) ! g(n)(a) asx! a.Similarly, we can derive a formula for limits as x approaches innity by taking x = 1

y.

limx!1

f(x)

g(x)= lim

y!0f(1=y)

g(1=y)

= limy!0

(1=y2)f 0(1=y)(1=y2)g0(1=y)

= limx!1

f 0(x)g0(x)

References

[1] Elementary Analysis: The Theory of Calculus, K. A. Ross.

[2] Calculus, G. B. Thomas and R. L. Finney, Pearson .

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