Mathematics

Session

Functions, Limits and Continuity - 2

Session Objectives

Sandwich Theorem

Limits of Trigonometric Functions

Limits of Exponential Functions

Limits of Logarithmic Functions

Limit at Infinity

Theorem

Note: If f(x)<g(x) for all x, then we can not say that .

x a x alimf(x)<limg(x)

x a x af g limf(x) limg(x)

If f and g are real functions defined in an open interval containing a

such that and both exist. Then,x alim g(x)

x alim f x

Sandwich Theorem

x a x alim f(x) = lim h(x) = m

x athen lim g(x) = m

Let f, g, h be real functions defined in an open interval containing ‘a’ such that

in the neighborhood of ‘a’f(x) g(x) h(x)

Limits of Trigonometric Functions

2 lim cos =10

1 lim sin =00

3sinlim =1, where is measured in radians.

0

4tanlim =1

0

Example-1(i)

1

1= =

x 0

sin xEvaluate: lim

sin x

x 0

sin xWe have lim

sin x

x 0

sin x× x

x=limsin x

× xx

x 0

x 0

sin xlim

x= ×sin x

limx

Solution :

Example-1(ii)

x 0

1- cosmxEvaluate : lim

1- cosnx

x 0

1- cosmxWe have lim

1- cosnx

2

x 0 2

mx2sin

2=limnx

2sin2

2

x 0

2

2 2

x 0

mxsin

2limmx

m 2= ×n nx

sin2lim

nx2

22 2

2 2 2

1m m= × =

n n1

Solution :

Example-1(iii)

3x 0

tanx - sinxEvaluate : lim

sin x

3x 0

tanx - sinxWe have lim

sin x

3x 0

sinx- sinx

cosx= limsin x

3x 0

sin - cosx= lim

sin xcosx

x 1

2

2x 0

x2sin

2= limsin xcosx

Solution :

Solution Cont.

2

2

x 0

22

x 0 x 0

xsin

2xxlim ×

2 2

=sinx

lim ×x ×lim cosxx

2

x 0

2

x 0 x 0

xsin

2xlim2

1= ×

2 sinxlim ×lim cosx

x

2

2

11 1= × =

2 21 ×1

Example-1(iv)

2x 0

A +B B - A2sin x sin x

2 2=lim

x

x 0

A +B B - Asin x sin x

2 2A +B B - A= 2lim . . .

2 2A +B B - Ax x

2 2

2x 0

cosAx - cosBxEvaluate : lim

x

2x 0

cosAx - cosBxWe have lim

x

Solution :

Solution Cont.

x 0 x 0

A +B B - Asin x sin x

2 2A +B B - A= 2 lim lim

2 2 A +B B - Ax x

2 2

2 2 2 2B - A B - A

= 1 1 =2 2

0

sinlim = 1

Example-1(v)

2 2

6x 0

sinx 1 - cosxEvaluate : lim

x

2 2

6x 0

sinx 1 - cosxWe have lim

x2

22

2 4x 0 x 0

x2sin

sinx 2=lim ×limx x

22

2

2 2x 0 x 0 x 0

xsin

sinx 1 sinx2=lim × lim lim2 xx x

2

1

21 1= 1 × × 1 =

2 2

Solution :

Example-1(vi)

-1

-11x

2

x - cos sin xEvaluate : lim

1 - tan sin x

-1

-11x

2

x - cos sin xWe have lim

1- tan sin x

-1Putting sin x = t x = sint

1x t

42

-1

-11x t

42

x - cos sin x sint - costlim = lim

1 - tant1 - tan sin x

Solution :

Solution Cont.

t4

sint - costcost - sint= lim

cost

t4

sint - cost= lim cos

- sint - costt

t4

1= - lim cos = -cos = -

4 2t

Limits of Exponential Functions

x

ex 0

a - 11 lim =log a

x

22

x e e

x 0 x 0

x1+x log a + log a + …- 1

a - 1 2!Proof : 1 lim = limx x

2

2e e

x 0

xx log a + log a + …

2!= limx

x

x 0

e - 12 lim =1

x

Limits of Exponential Functions (Cont.)

e=log a+0+0+0 …

e=log a

2 Substituting a= e in 1 , we get

x

ex 0

e - 1lim =log e =1

x

2e ex 0 x 0

x= lim log a+ lim log a + …

2!

Limits of Logarithmic Functions

x 0

log 1+xlim =1

x

x 0

log 1+xProof : lim

x

2 3

2

x 0 x 0

x xx - + - …

x x2 3= lim = lim 1- + - …x 2 3

2

x 0 x 0

x x=1- lim + lim - …

2 3

=1- 0+0 - 0 …=1

Example-2(i)x -x

x 0

e - eEvaluate : lim

sinx

x -x

x 0

e - eWe have lim

sinx

xx

x 0

1e -

e= limsinx

2x

xx 0

e - 1= lim

e sinx

Solution :

Solution Cont.

2x

xx 0 x 0

x 0

e - 1 2 1= lim × lim ×

2x sinxe limx

2 1= 1× × = 2

1 1

2x

xx 0

e - 1 2x= lim ×

2x e sinx

Example-2(ii)

h 0

log(e+h) - loge=lim

h

x e

logx - 1Evaluate : lim

x - e

x e

logx - 1We have lim

x - e

h 0

log(e+h) -1=lim Putting x=e+h, x e h 0

(e+h) - e

Solution :

Solution Cont.

h 0

hlog 1+

1e=lim ×

h ee

1 1=1× =

e e

h 0

hlog 1+

e=lim

h

x 0

log 1+xlim =1

x

Example-2(iii)

x 0

log a+ x - log a - xEvaluate : lim

x

x 0

log a+ x - log a - xWe have lim

x

x 0

x xlog a 1 + - log a 1 -

a alim

x

x 0

x xloga+log 1+ - loga- log 1-

a a=lim

x

Solution :

Solution Cont.

x 0 x 0

x xlog 1+ log 1-

1 1a a=lim × - lim × -

x xa a-a a

x 0 x 0

x xlog 1+ log 1-

a a=lim - lim

x x

1 1 2= 1× +1× =

a a a

Example -2 (iv)

x 0

1 + x - 1Evaluate : lim

log 1 + x

x 0

1+ x - 1We have lim

log 1+ x

x 0

1 + x - 1 1+ x +1= lim Rationalizing the numerator

log 1+ x 1 + x +1

x 0

1 + x - 1 1= lim

log 1 + x 1 + x +1

Solution :

Solution Cont.

x 0

x 0

1 1= lim

log 1 + x 1 + x +1lim

x

1 1 1= × =

1 1 +1 2

x 0

x 1= lim

log 1 + x 1 + x +1

Solution-2 (v)

mx nx

x 0

a - bEvaluate : lim

x

mx nx

x 0

a - bWe have lim

x

mx nx

x 0

a - 1 - b= lim

x1

mx nx

x 0

a -1 - b -1= lim

x

Solution :

Solution Cont.

x xm n

x 0 x 0

a -1 b -1=lim - lim

x x

xm n

x 0

a -1=loga - logb lim =loga

x

m

n

a=log

b

Thank you