Date post: | 01-Jan-2016 |
Category: |
Documents |
Upload: | merrill-bryan |
View: | 33 times |
Download: | 0 times |
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 :
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