differential calculus for jee main and …vkrclasses.com/downloads/pdf/limits-notes-vkr.pdf · 1.28...

1.26 DIFFERENTIAL CALCULUS FOR JEE MAIN AND ADVANCED

(v)x 0lim (sin x)x (vi)

x 0lim (n x)x

(vii)x 0lim

1x

1 sin x (viii)x 0lim (1)1/x

2. Suppose that limx a f(x) = and limx a g(x) = c,where c is a real number. Prove each statement.

(a) axlim [f(x) + g(x)] =

(b) axlim [f(x) g(x)] = if c > 0

(c) axlim [f(x) g(x)] = – if c < 0

3. Let 0)x(flimax

with f(x) 0 for x a,

b)x(glimax

0. Prove that )x(f

)x(glim

ax

4. Prove that the following limits donot exist :

(i) x2

lim

xtan

xtan

2 (ii) x 0lim

x1

)2(

(iii) x2

lim

11

xtan

xtan

e

e(iv) x 0

lim tan–1

x1

5. Evaluate the following limits :

(i) x 1lim

2xtan

xsin 1

(ii) x 0lim

1ln | x |

(iii) xlim x

2lnx

ee

(iv)

21

2

01 x

–

xexlim

6. Show that

x

x e

xlim .

7. If 4x

lim 2x

5)x(f

= 1, find

4xlim

f(x).

8. If 22 x

)x(flim

x = 1, find

(i)2x

lim

f(x) and (ii)2x

lim x

)x(f

C

9. Find axlim

x|x–a|2

10. Find the following limits by inspection.

(a)xln

xlim

0x

(b) xx e

xlim

3

(c)x ( /2)

lim (cos x)tan x

(d) 0xlim (ln x) cot x

(e)

xln

x

1lim

0x

(f) xlim (x + x3)

11. Evaluate

3x2 x 1

2x

3x 1lim2x 1

12. If f(x) = [x2 + 1][x + 1], where [.] denotes the greatest

integer function, find 1x

lim f(x).

METHODS OF EVALUATINGLIMITS

1.6 FACTORISATION ANDCANCELLATION OFCOMMON FACTORS

If f(x) and g(x) are polynomials such that f(a) = g(a) =0, then (x – a) is a factor of both f(x) and g(x). Now to

solve )x(g

)x(flim

ax , we cancel the common factor (x – a)from both the numerator and denominator, and againput x = a in the given expression. If we get ameaningful number, then the number is the limit of thegiven expression otherwise we repeat the process tillwe get rid of the indeterminate form (0/0).

Example 1. Find 9x

3xlim

23x

.

Solution Here, the denominator tends to zero asx 3 and the numerator also tends to zero. But sincex2 – 9 = (x – 3) (x + 3), we have

LIMITS 1.27

2x 3 x 3 x 3

x 3 x 3 1 1lim lim lim(x 3)(x 3) x 3 6x 9

In the solution of this problem we cancel x – 3, andone may think that this is illegitimate since x 3, andthe division by zero is not allowed. But this is not

the case here : the functions 9x

3xy

2

and

3x

1y

coincide identically for all x 3, and the

definition of a limit of a function for x a, does notinvolve the value of that function at the point x = aitself, and therefore the limits of the above functionsas x 3 are equal to each other. The essence of thistransformation is that the limit of the new function isfound easier than that of the original function.

Example 2. Evaluate3x

6xxlim2

3–x

Solution We cannot apply direct substitutionbecause the limit of the denominator is 0.

2x 63

x

x

+ -+

6

2

x 3

x 3

lim x + x – 3 = 0

lim x + 3 = 0

x xx

2 + - 6+3

6

2

x 3

x 3

lim x + x – 6 = 0

lim x + 3 = 0

Direct substitution fails here.Because the limit of the numerator is also 0, thenumerator and denominator have a common factorof (x + 3). Thus, for all x – 3, we can cancel thisfactor to obtain

x xx

2 + - 6+3

= (x 3)(x 2)

x 3

= x – 2, x – 3

It follows that

3x6xxlim

2

3–x

= .5–)2x(lim

3–x

Although correct, the second equality in the precedingcomputation needs some justification, since cancellingthe factor x + 3 alters the function by expanding itsdomain. However, the two functions are identical, except atx = – 3. From our previous discussions, we know that thisdifference has no effect on the limit as x approaches – 3.

Example 3. Find x 2lim

6

3

x 24x 16

x 2x 12

Solution x 2lim

6

3

x 24x 16

x 2x 12

= x 2lim

5 4 3 2

2

(x 2)(x 2x 4x 8x 16x 8)

(x 2x 6)(x 2)

= 168

14 = 12

Substitution for Limits x a

Suppose that x alim f(x) =

Let x = a + t, then f(x) = f(a + t).If x a then t 0 and f(a + t) , and we write

x alimf (a t)

= .

Example 4. Find 2xlim

x4

cos

4x2

.

Solution We put x – 2 = z, i.e. x = z + 2, and asx 2 we have z 0.

2xlim x

4cos

4x2

=

)2z(4

cos

)4z(zlim

0z

=

z4

sin–

)4z(zlim

0z

=z

4sin–

)4z(zlim

0z

= – 4

0z

lim z

4sin

z4

. 0zlim (z + 4) = –

16 .

1. xn – an = (x – a)(xn–1 + xn–2a + xn–3a2 + xn–4a3 +...+ xan–2 + an–1)

where n is even or odd positive integer.2. xn + an = (x + a)(xn–1–xn–2a + xn–3a2 – xn–4a3 + ...

+ (–1)n–1an–1)where n is odd positive integer. This formula is notapplicable when n is even.

Example 5. Find x

1)x1(lim

5 3

0x

.

Solution Let us put 1 + x = y5. As x 0, y 1.Then we have

x

1)x1(lim

5 3

0x

= 1y

1ylim

5

3

1y

5

3

1yyy

1yylim

34

2

1y


Example 6. Find 1xxx

1xxxlim

23

23

1x

Solution Here we have an indeterminacy of theform 0/0. Let us factorize the numerator and thedenominator of the function

1xxx

1xxxlim

23

23

1x

)1x()1x(x

)1x()1x(xlim

2

2

1x

2

2x 1

(x 1) (x 1)lim(x 1) (x 1)

02

0

1x

1xlim

1x

.

Example 7. Find 3

3 2x 10

x 1000lim

x 20x 100x

Solution This is also an indeterminacy of the form0/0. We have

x100x20x

1000xlim

23

3

10x

= 2

2

10x )10x(x

)100x10x()10x(lim

= )10x(x

100x10xlim

2

10x

The numerator of the fraction tends to 300 and thedenominator tends to zero. Consequently, the fractionin question is an infinitely large quantity and

3

3 2x 10

x 1000lim

x 2x 100x

does not exist.

Example 8. Find3–x

lim 6

332

23

xx

xxx

Solution At the point x = – 3 both the numeratorand the denominator turn into zero.We have x2 + x – 6 = (x + 3) (x – 2) andx3 + 3x2 – x – 3 = (x + 3) (x2 – 1), and hence, oncancelling the factor (x + 3), we obtain

3–xlim 6xx

3xx3x2

23

= 3–xlim 2x

1x2

=

2–3–1–)3(– 2

= – 58

In the same manner we can find

2xlim 6xx

3xx3x2

23

dne, 1xlim 6xx

3xx3x2

23

= 0.

Example 9. Evaluate

x 2lim 3 2

1 2 (2x 3)

x 2 x 3x 2x

Solution We have

x 2lim 3 2

1 2 (2x 3)

x 2 x 3x 2x

= x 2lim

1 2 (2x 3)

x 2 x(x 1)(x 2)

= x 2lim

x(x 1) 2(2x 3)

x(x 1)(x 2)

= x 2lim

2x 5x 6

x(x 1)(x 2)

=x 2lim

(x 2)(x 3)

x(x 1)(x 2)

= x 2lim

x 3

x(x 1)

= – 1

2

Example 10. Find y 2

1 1 1lim

y 2 x y 2 x

Solutiony 2

x x 2 y 1lim ·

x(x y 2) y 2

= – y 2

1lim

x(x y 2) = – 2x1

Here, y is a variable, so that it might be thought that weare dealing with functions of two variables. However,the fact that x as a variable plays no role in this problem;for the moment, x can be considered a constant.

Example 11. Evaluate x 3 x

x/2 1 xx 2

2 2 6lim

2 2

Solutionx 3 x

x/2 1 xx 2

2 2 6lim

2 2

0

form0

= 2x x

x/2x 2

2 8 6.2lim

2 2

= x x

x/2x 2

(2 2)(2 4)lim

2 2

= x x/2 x/2

x/2x 2

(2 2)(2 2)(2 2)lim

2 2

= 2 × 4 = 8.

Example 12. Evaluate 3 2

2x 1

x x log x log x 1lim

x 1

Solution The given limit= 3 2

2x 1

x 1 x 1 logxlim

x 1

LIMITS 1.29

=

2

1

x –1 x + x +1 – x –1 x +1 logxlim

x -1 x +1x

=

2

1

x -1 x + x +1- x +1 logxlim

x -1 x +1

x

= ( )

( )21 +1+1- 1+1 log1 3=

1+1 2 .

Example 13. If the limit2

2x 2

3x ax a 3lim

x x 2

exists, find a and the limit.Solution We see that the denominator 0. For

the limit to exist, we must have the 0

0 form. Hence

the numerator must 0, i.e.

x 2lim 3x2 + ax + a + 3 = 0

12 – 2a + a + 3 = 0 a = 15.

Now 2

2x 2

3x ax a 3lim

x x 2

= 2

2x 2

2x 15x 18lim

x x 2

=x 2

3(x 2)(x 3)lim 1

(x 2)(x 1)

.

Example 14. If 2x 0

f (x)lim 2x

then evaluate the

following limits, giving explicit reasoning.

(i) x 0lim f (x)

, (ii)x 0

f (x)limx

where [ . ]

denotes the greatest integer function.

Solution (i) Let l = x 0lim f (x)

= 2

2x 0

f (x)lim ·xx

Now argument of G.I.F is tending towards zero and

from positive side as 2x 0

f (x)lim 2x

and x2 0+

we have 22

f (x) ·xx

0+ l = 0.

(ii) We write x 0

f (x)limx

= 2x 0

f (x)lim ·xx

and assume A = 2f (x) ·xx

Now x 0+ A 0+

and x 0– A 0–

x 0lim [A]

= 0 and –x 0

lim [A]

= – 1

The given limit does not exist.Example 15. Discuss the behaviour of

(x) =

n1n1

n0

kmk

1m1

m0

xb....xbxb

xa.....xaxa

where a0 0, b0 0 as x tends to 0 by positiveor negative values.Solution If m > n,

0xlim (x) = 0.

If m = n, 0x

lim (x) = a0/b0.

If m < n and n – m is even, (x) or (x) –according as a0/b0 > 0 or a0/b0 < 0.If m < n and n – m is odd, then (x) as x 0+

and (x) – as x 0–, or (x) – as x 0and (x) as x 0–, according as a0/b0 > 0 ora0/b0 < 0.

E1. Evaluate the following limits :

(i)x 2lim

3 2

3

x 3x 9x 2

x x 6

(ii)2

2x 0

(x 1)lim

(x 1) 1

(iii)h

x)hx(lim

22

0h

(iv) 5t 1

1 tlim

1 t


(i)y 1

| y 1 | y 1l im

| y 1 | y 1

(ii)3

x 3

x 27lim| x 3 | (x 3)

3. Let f(x) = 32

22

2

–xx

xx

find

(a) )x(flim2–x (b) )x(flim

1x

(c) )x(flim3–x (d) )x(flim

0x


4. Evaluate 1x

1]x[itlim

2

2

1x

where [.] denotes the

greatest integer function.

5. Evaluate x 5lim ]x[x

xx

2092

where [.] denotes

the greatest integer function.

6. Evaluate the following limits where [.] denotesthe greatest integer function :

(i)x 2

x [x]limx 2

(ii)

x 2

x [x]limx 2

(iii)2 2

2x 1

[x ] [x]limx 1

(iv) 3 3

x 10

[x ] xlim[x] x

D7. Evaluate the following limits :

(i)4

5x 1

x 3x 2limx 4x 3

(ii) 22

23

1x )1x(

2x5x4xlim

(iii) 2312

23

57

1

xx

xxlimx

(iv)33

27623

24

3

xxx

xxlim

–x


(i)2

4t 1

t t tlim

t 1 t 1

(ii)1xsin3–xsin2

1–xsinxsin2lim

2

2

6/x

(iii)xcotxcot2

xcot1lim

3

3

4/x

9. Find a number c so that

12x7x

12x5cxxlim

2

23

3x

exists. Also find the

corresponding limit.10. Evaluate the one-sided limits in the following :

(i) x 2lim

2 2x [x ]

(1 | x 2 |)

(ii) x 0lim

[x] |x|x(e 2)

[x] | x |

where [.] denotes the greatest integerfunction.

1.7 RATIONALIZATIONIf in any limit, the denominator or numerator involvesthe radical sign then we can rationalize the irrationalexpression by multiplying with their conjugates toremove the indeterminacy.

Example 1. Evaluate x 1lim

4 15x 1

2 3x 1.

Solution x 1lim

4 15x 1

2 3x 1

= x 1lim

(4 15x 1)(2 3x 1)(4 15x 1)

(2 3x 1)(4 15x 1)(2 3x 1)

= x 1lim

(15 5x)

(3 3x)

×

2 3x 1

4 15x 1 =

5

6

Example 2. Evaluate the following limits :

(i)x 0lim

1 x 1 x

x

(ii)x 1lim

2

(2x 3) x 1

2x x 3

Solution (i) The given limit takes the form 0

0when x 0. Rationalizing the numerator, we get

x 0lim

1 x 1 x

x

= x 0lim

1 x 1 x 1 x 1 x

x 1 x 1 x

= x 0lim

(1 x) (1 x)

x 1 x 1 x

LIMITS 1.31

= x 0lim

2x

x 1 x 1 x

= x 0lim

2

1 x 1 x

= 2

2 = 1.

(ii) We have x 1lim

2

(2x 3) x 1

2x x 3

= x 1lim

(2x 3) x 1

(2x 3)(x 1)

= x 1lim

(2x 3) x 1

(2x 3) x 1 x 1

= x 1lim

2x 3

(2x 3) x 1

= 1

(5)(2)

=

1

10

.

Example 3. Evaluate 1

x 1

1 xlim sin1 x

Solution1

x 1

1 xlim sin1 x

= sin–1x 1

1 xlim

1 x

= sin–1

)x)(x(

xlimx 11

11

= sin–1

x 1

1lim

1 x

= sin–1

621 .

Example 4. Evaluate 2x 2

x 2lim

x 4 x 2

Solution 2x 2

x 2lim

x 4 x 2

= 2

2 2x 2

x 2 ( x 4 x 2).limx 4 x 2 ( x 4 x 2)

= 2

2x 2

(x 2)( x 4 x 2)lim

(x 4) (x 2)

= 2

2x 2

(x 2)( x 4 x 2)lim

x x 2

=2

x 2

(x 2)( x 4 x 2)lim

(x 2)(x 1)

=

2

x 2

x 4 x 2lim

x 1 = 0.

1 1 n 1 n 2 1 n 3 2 n 1n n n n n n n n(x a )(x x a x a ..... a ) (x a)

1 1 n 1 n 2 1 n 3 2n n n n n n n(x a )(x x a x a ...) (x a)if nisodd

Example 5. Evaluate )x(

)x(limx 1

273

1

Solution 3 (7 x) – 2

= 427787

3132

.)x()x(

)x(//

= 2/3 1/3

(x 1)

(x 1) (x 1) .2 4

...(1)

3

x 1

(7 x) 2lim

(x 1)

= – 2/3 1/3x 1

(x 1)lim

(x 1)(7 x) (7 x) .2 4

[from (1)]

= 42771

31321 .)x()x(lim

//x

= 1

4 2 4 = –

1

10.

Example 6. Calculate 2

33 2

1x )1x(

1x2xlim

Solution We substitute 3 x = tThen, for the variable t, the expression under the limitsign can be written in the form

23

2

)1t(

1t2t

The number to which the new variable t tends, asx 1, can be found as the limit of the function

t(x) = 3 x as x 1,

i.e. 1xlim)x(tlim 3

1x1x


Thus we have

222

2

1t23

2

1t )1tt()1t(

)1t(lim

)1t(

1t2tlim

91

)1tt(1lim

221t

.

Example 7. Evaluate

2 2x 2a

x 2a x 2alim

x 4a

Solution

2 2x 2a

x 2a x 2alim

x 4a

0form

0

=

2 2 2 2x 2a x 2a

x 2a x 2alim lim

x 4a x 4a

= 2 2x 2a

( x 2a )( x 2a )lim

x 4a ( x 2a )

+ x 2a

( x 2a )lim

( x 2a )( x 2a )

=

x 2a

x 2a 1lim

(x 2a)(x 2a).2 2a 4a

=

x 2a

x 2a 1lim

x 2a 2 2a 2 a

= 0 + 1 1

2 a 2 a.

FFFFF1. Calculate the value of the function

f(x) = 1–1x

x

at several points near x = 0 and hence estimatethe limit of f(x) as x 0.


(i) x 0lim

x5 x 5 x

(ii) x 9lim

3 x4 2x 2

(iii) x 1lim

3 32x x

(x 1)

(iv) x 1lim 1x2

2x5

3. Evaluate 1x

2xx4lim

2

1x

4. Evaluate x 1lim 2

x 1

6x 3 3x

5. Find numbers a and b so that

1x

1baxlim

0x

.

EEEEE6. Evaluate a

x 3

x 3lim logx 6 3


(i)x2x

xx27xx1lim

2

22

2x

(ii) x 2lim

1 2 x 3

x 2

(iii) x 1lim 4

x 1

x 17 2

(iv) 3x 0

1 x – 1 xlim

1 x – 1 x

8. Evaluate 3x4x

6x2x6x2xlim

2

22

3x

.

9. Evaluate x 2lim 3 3

x 7 3 2x 3

x 6 2 3x 5


(i) 22

22

1 53

108

xx

xxlimx

LIMITS 1.33

(ii) 3

4x 0

8 3x 2lim

16 5x 2

(iii) 111

113 2

5 3

0

)x()x(

)xlimx .

1.8 LIMIT USING EXPANSIONSERIES OF FUNCTIONS

In this method basically we use the series expansionof sin x, cos x, tan x, log(1+x), ax, ex, etc. to evaluatethe limit. Following are some of the frequently usedseries expansions:

(i)2 2 3 3

x x1na x 1n a x 1n aa 1 .........a 01! 2! 3!

(ii)2 3

x x x xe 1 ............1! 2! 3!

(iii) ln (1+x) = 2 3 4x x x

x .........for 1 x 12 3 4

(iv)3 5 7x x xsin x x .......

3! 5! 7!

(v)2 4 6x x xcosx 1 ......

2! 4! 6!

(vi) tan x = 3 5x 2x

x ........3 15

(vii) tan–1x = 3 5 7x x x

x .......3 5 7

(viii) sin-1x = 2 2 2 2 2 2

3 5 71 1 .3 1 .3 .5x x x x .......

3! 5! 7!

(ix) sec x = 2 4 6x 5x 61x

1 ......2! 4! 6!

(x) (1 + x)n = 1 +2nx x

n(n 1)1 2 +.... for –1< x< 1,

n Q.

(xi) (1 + x)1/x =

.....xxe 2

2411

211

Using the above expansions, we canfind other expansion series. For example to find the

expansion series of sin2x, we write, sin2x = 2

x2cos1

and use the expansion series of cos x with x replacedby 2x.

Maclaurin's Theorem

f(x) = f(0) + xf'(0) + ...)0("'f!3

x)0("f

!2

x 32

........)(f!n

x........ n

n

0

Example 1. Expand sin x in powers of x.Solution Here f(x) = sin x, Hence f(0) = 0,

f'(x) = cos x, f'(0) = 1,f"(x) = – sin x, f"(0)= 0,f'"(x) = – cos x, f"'(0) = – 1

.............

fn(x) = sin

2

nx fn(0) = sin

2

n

Thus sin x = x – ...!n

2n

sinx....–

!5

x

!3

xn

53

Example 2. Expand ln(cos x) in powers of x.Solution Here f(x) = ln(cos x),

f'(x) = – tan x = – t, say,f"(x) = – sec2 x = – (1 + t2),f'"(x) = – 2 tan x sec2 x = – 2t (1 + t2),f(4)(x) = – 2 (1 + 3t2) (1 + t2) = – 2 (1 + 4t2 + 3t4),f(5)(x) = – 2 (8t + 12t3) (1 + t2)

= – 2(8t + 20t3 + 12t5),f(6)(x) = – 2 (8 + 60t2 + 60t4) (1 + t2)

= – 2(8 + 68t2 + 120t4 + 60t6)Hence,

f(0) = 0, andf'(0) = f(3)(0) = f(5)(0) = ...= 0, alsof"(0) = – 1, f(4)(0) = –2, f(6)(0) = – 16.

Hence ln(cos x) = – .............!

x–

!

x–

!

x

616

42

2

642


x

2

e 1 x

x

Solution x 0lim

x

2

e 1 x

x

= x 0lim

2

2

x1 x ....... 1 x

2!

x

= x 0lim

21 x x ............2! 3! 4! =

1

2.



tan x sin x

x

Solution x 0lim 3

tan x sin x

x

= x 0lim

3 3

3

x xx ........ x .......

3 3!

x

x 0lim

3

3

1 1 x ........3 3!

x

=

1

3 +

1

6 =

1

2.

Example 5. Evaluate xxtane

xlne

lim

x

x

1

0

Solutionxxtane

xlne

lim

x

x

1

0

x

x 0

e ln(1 x) ln elim

tan x x

[form 0/0]

= x

15x2

3x

x

1...3x

2x

x...!3

x!2

xx1

lim53

3232

0x

= 3

x 0 3

1x (1 termscontainingxand itshigher powers)

6lim1

x (1 termscontainingxand itshigher powers)3

= – 2

1.


5sinx 7sin2x 3sin3xlim

x sinx

Solutionxsinx

x3sin3x2sin7xsin5lim

20x

=

3 3 3

3x 0 2

x (2x) (3x)5 x .... 7 2x .... 3 3x ....

3! 3! 3!lim

xx x ....

3!

=

...

!3x

1x

!3x81

!3x56

!3x5

lim3

3

333

0x

= 5!3

81565

.


2xn(1 x) sin x

2x tan x sin x

Solution x 0lim

2xn(1 x) sin x

2x tan x sin x

=x 0lim

2 3 3 5 2

3

x x x x xx ..... x .....

2 3 3! 5! 2tan x sin x

x . .x x

= 1

3 +

1

6 =

1

2.


3 6

2 x 2x 0

cosx 1 n(1 x )lim

x (e 1 x )

Solution 2

3 6

2 x 2x 0

cos x 1 n(1 x )lim

x (e 1 x )

=

6 126

2 4x 02 2

x x1 ........ 1 x

2 2lim

x xx 1 ....... 1 x

1 2

= x 0

1.....

2lim 1

1.......

2

.

Example 9. Find sin x

2x 0

e sin x 1lim

x

Solutionsin x

2x 0

e sin x 1lim

x

=

2 3

2x 0

sin x sin x sin x1 ..... sin x 1

1 2 3lim

x

= 2

x 0

sin x 1 sin x 1 1lim ...

x 2 3 2 2

.

LIMITS 1.35


1/37 x) 2

x 1

Solution Put x 1 + h

h 0lim

1/3(8 h) 2

h

= h 0lim

1/3h2 . 1 2

8h

= h 0lim

21 1 h1

1 h 3 3 82 1 . ....... 13 8 1. 2

h

= h 0lim 2 ×

1

24 =

1

12.

Example 11. Find 7 5

3 2x 1

x 2x 1limx 3x 2

.

Solution This is of the form 0

0 if we put x = 1.

Therefore we put x = 1 + h and expand.

7 5

3 2x 1

x 2x 1limx 3x 2

= 7 5

3 2h 0

(1 h) 2(1 h) 1lim(1 h) 3(1 h) 2

= 2 2

2 2h 0

(1 7h 21h ...) 2(1 5h 19h ...) 1lim(1 3h 3h ...) 3(1 2h h ) 2

= 2

h 0

3h h ...lim3h ....

= h 0

3 h ...lim3 ....

= 1.

Example 12. If 4x 0

a cosx bxsin x 5lim

x

exists,

find a, b and the limit.Solution As x 0, x4 0 the limit must be in

0

0 form. Hence

x 0lim acos x + bx sin x – 5 = 0.

a – 5 = 0 a = 5.

Limit =

2 4 3

4x 0

x x x5 1 ....... bx x ...... 5

2 2 3lim

x

=

2 4

4x 0

5 5 bb x x ....

2 24 6limx

For limit to exist, x4 must cancell from the numerator.Hence we assume the coefficients of all powers of x

less than 4 to be zero 5

b 02

b = 5

2.

Now the limit =

4

4x 0

5 bx ........

24 6limx

= 5 b

24 6

Limit = 5 5 5

24 12 24

.

Example 13. If x x

x 0

Ae Bcosx Celim 2

xsin x

,

find A, B and C.Solution The given limit is equal to

2 2 2

3x 0

x x xA 1 x ....... B 1 ....... C 1 x .......

2 2 2lim

xx x .......

3

= 2

2x 0 2

A B C(A B C) (A C)x x

2limx

x 1 ......3

For limit to exist, x2 must cancel from the numerator.Hence we assume A – B + C = 0 ...(1)

and A – C = 0 ...(2)

Limit = A B C

22

. ...(3)

On solving these three equations, we get A = 1, B= 2, C = 1.Example 14. Show that

1/x

2x 0

1(1 x) e ex 11e2lim

24x

.

Solution Let y = (1 + x)1/x

ln y = )x1ln(x

1


=

...x

3

1x

2

1x

x

1 32

= .....x3

1x

2

11 2 ,

Now ......xx

ey

2

21

211

= e.......xx

e 2

21

211

= 21 1e 1 x x ...... .....

2 3

+ 2

21 1 1x x ...... ....

2 2 3

=

.....x

8

1

3

1x

2

11e 2

=

.....xxe 2

2411

211

Now 2

x/1

0x x

ex21

e)x1(lim

2

2x 0

1 11 1e(1 x x ....) e ex

2 24 2limx

=

24

e11.

Example 15. Let f(x) be a function such that

x 0

f(x)lim

x = 1. Find the values of a and b such that

1130

)}x(f{

xsinb)xcosa(xlimx .

Solution Since, 3x 0

x(1 a cosx) bsin xlim 1

{f(x)}

3

5342

0

534211

)}x(f{

....!

x!

xxb

!x

!x

ax

limx

= 1

3 5

3x 0

a b a bx(1 a b) x x ...

2! 3! 4! 5!lim{f(x)}

= 1

2

2

3x 0

(1 a b) a b a bx ...

2! 3! 4! 5!xlimf(x)

x

= 1

Since the limit exists we must have 1 + a –b = 0 and

– a b

2! 3! = 1 –3a + b = 6

Solving these, we get a = –5/2 and b = –3/2.

GGGGG1. Evaluate the following limits :

(i) 2

3

0x x

x31

1x1lim

(ii))x1ln(xx

xxxsinelim

2

2x

0x

(iii)3

2

0x x2

x1xcoslim

(iv) 20x x

)x21ln(x2lim


(i)xx

xxx 1

1

0 tan2sin2lim

(ii) x 0lim

3 sin x sin 3x

x sin x

(iii) 0xlim

x

3

e sin x x(1 x)

x

(iv) 0xlim

3

2

1 3x 1 2x

x

3. Evaluate 5

3

0x x

xx3xtan3lim

4. Evaluate 1 1

3x 0

sin x tan xlim

x

5. Evaluate 3

x

0x x

)2xe)(x2(lim

6. For what values of constants C and D is it true

that 2)DCxx7sinx(lim 23

0x

LIMITS 1.37


(i)x 0lim

2

3

(1 x ) (1 x)

(1 x ) (1 x)

(ii)x 0lim

3 42 2

2

1 x 1 x

x x

(iii)x501)x1(

1x31lim

50

3

0x

(iv) 3

22

0x x

x)x1(lnlim


(i)x

bx1ax1lim

nm

0x

(m, n N)

(ii)x

bxaxlim

nm

x

1110

(m, n N)

(iii)

2x

11

x91x21lim

43

0x

(iv)xxtan

e4x14lim

1

x4 2

0x

2


(i)7

53

0x x8x

2x

)xcos1(xsinlim

(ii) 7

53

0x )x(sin

120/x6/xxxsinlim

(iii))xsin(

x1elim

2

xcosx

0x

(iv)3

213

02

1

x

xsin–)xln(xcosx

lim

–

x

10. Evaluate 5x 0

1 1 x2sin x ln – 3x

2 1 xlimx

11. Evaluate

4x 33 24 2

7x 0

x e – sin xlim

x

12. Find the values of a and b so that

x 0lim 4

(1 axsin x) bcos x

x

may tend to a definite limit and also find limit.13. For what values of the constants a, b is ,

x 0lim 3 2

sin3x ab

x x = 0 ?

14. If 3x 0

1 1 1 axlim1 bxx 1 x

exists and has

the value equal to , then find the value of

1 2 3

a b .

FFFFF

1.9 STANDARD LIMITSFollowing are some basic limits which are used fre-quently in solving the limits.

(i) 0xlim

sin xx = 1 = 0x

lim

tan x

x

= 0xlim x

xtan 1

= 0xlim

1sin x

x

[ where x is measured in radians ]

Also, 0xlim 2

1 cos xx

= 0x

lim 2

(1 cosx)(1 cosx)x (1 cosx)

= 0xlim

2

2(sin x)

x (1 cos x) = 0xlim

2sin x 1x (1 cos x)

= 21

.

(ii) 0xlim

xa 1

x

= ln a (a > 0).

In particular 0xlim

xe 1

x

= 1

(iii)x 0lim x

)xln( 1 = 1


x 0lim

alog (1 x)x

= loga e, ( a > 0, a 1)

(iv)x

)x(lim

n

x

110

= n

axlim

n nn 1x a n a

x a

.

(i) These limits help in finding limits in 0/0 form andmost of them are based on x approaching 0. Incase, if x approaches a, we use the substitutionx = a + t so that t approaches 0 when x approaches a.

(ii) These formulae are also applicable when afunction, say f(x) has been replaced for x. But itmust be ensured that as x approaches a, f(x) mustapproach 0.

For instance, x alim

sin f (x)f (x) = 1, if f(x)

approaches 0 as x approaches a.

But x alim

sin f (x)g(x) where f(x) and g(x) approach 0 as

x approaches a, should not be immediately taken as 1.

For example, x 0lim

sin 2x

x =

x 0lim

sin 2x

2x. 2 = 2

Further, x alim

sin f (x)f (x) should not be taken as 1 when

sin f(x) approaches 0 as x approaches a, but f(x) itselfdoes not approach 0.

For example, x 0lim

sin( x)x

1, since – x does notapproach 0 as x approaches 0. In fact, the limit is equalto 0.

Why Radian Measure is Used

Throughout calculus, angles are measured in radians, asis customary in calculus. If we measured angles in degreesinstead, the formulas for the limits of the trigonometricfunctions would be more complicated. Each formulawould have an extra factor, /180, as we will now show.Earlier, it was shown that when angles are measured

in radians, 1sin

lim0

.

When angles are measured in degrees, this limitis not 1. Let sin denote the sine of an angle of degrees. The following table suggests that the limitis much smaller (angles measured in degrees; data tofour significant figures):

10 5 1 0.1sin 0.1736 0.08716 0.01745 0.001745

sinθθ

0.01736 0.01743 0.01745 0.01745

The data suggest that 0

sinlim

is about0.01745. We can find that the limit is precisely /180.

Example 1.x

xtanlimx

0

Solution radianx180

x

Now x 0 x 0

xtantan x 180lim lim

x x

180

180180

180

0

x

x.

x

xtan

limx

.

Example 2. Evaluate 0x

lim

3xe 1

x / 2

Solution 0xlim

3xe 1

x / 2

= 0x

lim 2 × 3

3xe 1

3x

= – 6.

Example 3. Compute x 0lim

sin 2x

sin3x

Solution We have x 0lim

sin 2x

sin3x

= 0xlim

sin 2x 2x 3x. .

2x 3x sin3x

= 2x 0

sin 2xlim

2x

.

2

3 .

3x 0

3xlim

sin 2x

, x 0

= 1 . 2

3 .

1

3x 0

sin3xlim

3x

= 2

3 × 1 =

2

3.

LIMITS 1.39


3

2

x 8

x 4

Solution The given expression is of the form3 3

2 2

x (2)

x (2)

= 3 3x (2)

x 2

/ 2 2x (2)

x 2

x 2lim

3

2

x 8

x 4

= x 2lim

3 3x (2)

x 2

/ x 2lim

2 2x (2)

x 2

= 3(22)/(2(21)) (using x alim

n nx a

x a

= nan–1 )

= 12 /4 = 3.

Example 5. Find 5/1

3/1

1x x1x1lim

.

Solution We have 4

5/1

55/1

1x)1(5

)1(x)1(xlim

= 5

and)1(x

)1()x(lim 3/1

33/1

1x

= 3 (–1)2 = 3

Hence the limit = )1(x)1(xlim 5/1

3/1

1x

= 3

5.

Example 6. Evaluatex h x h x

h 0

a a 2alim

h

, a > 0

Solution Limit = h 0lim ax

h h

2

a a 2

h

= 2h h

x2h 0

a 2a 1lim a

h

= 22

x a 1a

h

= ax ln2a.

Example 7. Evaluate x a

x a

a alimx a

, a > 0.

Solution Put x = a + h

Limit = a h a

h 0

a alimh

= h 0

lim

h)1a(a ha

aa lna

Example 8. Evaluate

x 0

sin xlim

x .

Solution Since, sin x

0 1x

, sin x

x = 0 in the

neighbourhood of x = 0.

Hence

x 0

sin xlim

x = 0.

With a similar reason, 1

x 0

tan xlim 0

x

.

Example 9. Evaluate

x 0

tan xlim

x .

Solution Since, tan x

1 2x

in the neighbourhood

of x = 0,

ta n x

x = 1 . Hence,

x 0

tan xlim

x = 1.

With a similar reason, 1

x 0

sin xlim 1

x

.

Example 10. Evaluate

xtan

x2lim

0x, where [.]

denotes greatest integer function.Solution We know when x 0

1xtan

x

x1

tan x

–2 < 2x

1tan x

So,

xtan

xlimx

20 = –2.

Example 11. Evaluate ex1xlnlim

ex

.

Solution Limit =

1

exe

e/xlnlimex

= t 1

ln tlimt 1 e (putting x/e = t)

Put t = 1 + y =

y 0

ln 1 y 1limey e


Example 12. Evaluate 0xlim 3

tan x sin x

x

Solution 0xlim 3

tan x sin x

x

= 0xlim 3

tan x(1 cos x)

x

= 0xlim

2

3

xtan x . 2sin

2x

= 0xlim

tan x

x .

2xsin

2x2

= 1.


2

0x x

)xcossin(lim

Solution Here cos2x as x 0.We change cos2x to sin2x, so that it tends to 0.

2

2

0x x

)xcossin(lim

= 2

2

0x x

)}xsin1(sin{lim

= 2

2

0x x

)xsinsin(lim

= 2

2

0x x

)xsinsin(lim

=

2

2

2

2

0x x

xsin

1xsin

)xsinsin(lim

= 2

2

0x2

2

0x x

xsinlim

xsin

)xsinsin(lim

= 1 × × 1 =

Example 14. Evaluate 1xlim

x sin{x}x 1- , where {x}

denotes the fractional part of x.Solution As x 1– , {x} can be replaced by 1–

h, where h is a small positive quantity.As x 1+ , {x} can be replaced by h, where h is asmall positive quantity.

x 1 h 0

x sin{x} 1 hlim lim sin(1 h)x 1 h

= –

x 1 h 0

x sin{x} sin hlim lim(1 h)x 1 h+ +® ®

= +-

= 1 × 1 = 1Since, L.H.L. R. H.L. the limit does not exist.

Example 15. Evaluate x ax a

cosx ln(x a)lim

ln(e e )

Solution x ax a

cosx ln(x a)lim

ln(e e )

Put x = a + h

= cos a · a h ah 0

ln hlim

ln(e e )

= cos a · hh 0 a

ln hlim

e 1ln e · ·h

h

= cos a · hh 0

ln hlim

e 1a ln ln h

h

= cos a · hh 0

1lim

a ln(e h)1

ln h ln h

= cos a.

Example 16. Evaluate 0xlim x

x2cos1

Solution f(x) = x

x2cos1

= x

|xsin|2

x

xsin2 2

,

` but |sin x| =

0x2/–if,xsin–,2/x0if,xsin

Hence f(0–) = x 0 x 0

sin xlim f(x) lim – 2 – 2

x

and f(0+) = x 0 x 0

sin xlim f(x) lim 2 2

x

.

Example 17. Evaluate 4 2x 0

n cosxlim

1 x 1

.

Solution Limit =

2 1/4x 0

22

n(1 cosx 1)(cosx 1)lim

(1 x ) 1(cosx 1) x

x

LIMITS 1.41

= 1 12

1 24

.

Example 18. Evaluate 2t 0

n cos(sin t)lim

t

Solution 2t 0

n cos(sin t)limt

22

2t 0 2

sin t(sin t) 2sinn 1 2sin22lim .

sin t t2sin2

=

2

2

2t 0

sin tsin

sin t 12lim 2. .sin t 24.t

2

.

Example 19. Find the limiting value of

3xxsin2x2tan

as x tends to zero.

Solutionx 0lim 3

sin 2x 2sin xcos2x

x cos2x

= 0x

lim 3

2sin x[cosx cos2x]

x .1

= 2 x 0lim 2x

sinsin2 2x

2x3

= 2 x 0lim

2x32x3

2

sin3.sin x

x2

2 = 3.

Example 20. Let a = min [x2 + 2x + 3, x R] and

b = x xx 0

sin xcosxlim

e e , then find the value of

n

0r

rnrba .

Solution a = minimum value of (x + 1)2 + 2 a = 2

b = 2xx 0

sin 2xlim

2(e 1).2x2x

= 2

1.

Now

n

0r

rnrba =

rnr

212 =

n

0r

r2n 2

21

=

n

0r

rn 4

21

= n21

[1 + 4 + 42 + ...... + 4n]

= n21

314 1n

= n

1n

2.314

.

Example 21. Evaluate x

xx 0

e ln(x e)lim

e 1

.

Solution

x

xx 0

x(e ln e 1

elime 1

xx

= x

x 0

ln 1 (x e)(e 1)lim

x x

= 1 – x 0

ln 1 (x e)lim

xe·

e

= 1 – e1

.

Example 22. Evaluate 2 x xx 0

n(cos3x) 2sinxlim .

x e e

.

Solution

x

2 x x 2 2xx 0 x 0

n(cos3x).2sin x n(cos3x).2sin x.elim lim

x (e e ) x (e 1)

= x 0lim

n(1 cos3x 1)(cos3x 1)

(cos3x 1)

2x2

2sin x 1

x e 12x

2x

= x 0lim

2

2

3x2sin

n(1 cos3x 1) 2(cos3x 1) 2x


2x

2sin x 1

x e 1x

= 1 × (–1) × 9

4 × 2 × 1 = –

9

2.

Example 23. Evaluate 211x xcos

x1lim

.

Solution Put cos-1 x = y and x 1– y 0+

211x xcos

x1lim

= 20y y

ycos1lim

Now rationalizing the numerator, we get

= ycos1y

ycos1lim20y

= ycos1

1lim.

y

ycos1lim

0y20y

= 41

21.

21

.

Example 24. Evaluate

1 1

2x 1

sin cos xlim

1 x.

Solution

1 1 1

1 2x 1

sin cos x cos xlim

cos x 1 x

=

1 1

2 1x 1

cos x cos xlim ·

1 x cos x

= 1

2x 1

cos x1lim

2 1 x

= 1

2x 1

1 cos ( x)lim

2 1 x

put cos–1(–x) = x = – cos

=0

1lim

| sin |2

=0

1lim

sin2

= 21

.

Example 25. Solve 1

2x

2

tanx sin{tan (tanx)}lim

tanx cos (tanx)

Solution Here, RHL

= )x(tancosxtan

)}x(tansin{tanxtanlim

x2

1

2

= 2

x2

tan x sin(x )lim

tan x cos (tan x)

{ tan–1 (tan x) = x – , when x > 2

}

= 2

x2

sin x1

tan xlimcos (tan x)

1tan x

=

1 0

1 0

= 1.

Now, LHL

=

1

2x

2

tan x sin{tan (tan x)}lim

tan x cos (tan x)

= 2x

2

tan x sin(x)limtan x cos (tan x)

1as tan (tan x) x, when x2

= 2

x2

sin x1

tan xlimcos (tan x)

1tan x

= 1 0

1 0

= 1

1

2x

2

tan x sin{tan (tan x)}lim

tan x cos (tan x)

= 1.

LIMITS 1.43

Example 26. Evaluate h 0lim

x h x(x h) x

h

(x > 0).

Solution h 0lim

x h n (x h) x n xe e

h

= h 0lim

x n xe x h n (x h) x n xe 1

h

= xxs

s 0

e 1lim

s

h 0lim

x h n (x h) x n x

h

where s = x h n (x h) x n x

= xx .h 0lim

h

nxxxh

nnxhx

1

= xx h 0

( x h x)lim n x

h

x/h

h 0

(x h) hlim n 1

x x

= xx

xx

xn 12

.

Example 27. Evaluate x

xcos1lim

0x

Solutionx

xcos1lim

0x

= xcos1

1.

x

xcos1lim

0x

= xcos1

1.

)2/x(2

|)2/xsin(|2lim

0x

Now, we have

LHL = xcos1

1.

2/x

)2/xsin(.

2

1lim

0x

= 2

1

2

1.

2

1

.

and RHL = xcos1

1.

2/x

)2/xsin(.

2

1lim

0x

= 2

1

2

1.

2

1 .

Hence, limit does not exist.

Use of Substitution

Sometimes in solving limit problem we convert x alim f(x)

by substituting x = a + h or x = a – h as h 0lim f(a + h) or

h 0lim f(a – h) according to the need of the problem.

Example 28. Compute x 3lim

x 3e e

x 3

Solution Put y = x – 3. So, as x 3, y 0.

Thus x 3lim

x 3e e

x 3

= y 0lim

3 y 3e e

y

= y 0lim

3 y 3e . e e

y

= e3 y 0lim

ye 1

y

= e3 . 1 = e3 .

Example 29. Find x /4

ln tan xlim

1 cot x Solution Put x = t + /4

t 0

ln tan(t / 4)lim

1 cot(t / 4)

= t 0

1 tan tln

1 tan tlimcot t 1

1cot t 1

= t 0

ln(1 tan t)lim

2 tan t1 tan t

+ t 0

ln(1 tan t)lim

2 tan t1 tan t

= 1

2 [1.1 + 1.1] =

2

2 = 1.


Example 30. Evaluate 3/23/x )xcos21(

6xsin

lim

Solution Putting x – z3

= 2/3z 0

cos z2lim

(1 2cosz 3 sinz)

= 3/20z )zsin3zcos1(

zsinlim

= 3/23/20z

2z

cos32z

sin2z

sin2

2z

cos2z

sin2lim

=3/2

3/13/1

0z

2z

cos32z

sin2

2z

cos2z

sin2

lim

= 0

)3(

1.0.23/2

3/1

.

Example 31. Evaluate /4

(1 tan )lim

(1 2 sin )

Solution Let P = x 0lim

(1 tan )

(1 2 sin )

0form

0

Put = 4

+ h

P =

h 0

1 tan h4lim

1 2 sin h4

= h 0

1 tan h1

1 tan hlim

1 11 2 cosh sin h

2 2

0

form0

= h 0

2 tan hlim

(1 tan h)(1 cos h sin h)

= h 0

2 tan hlim

(1 cos h)(1 cos h)(1 tan h) sin h

(1 cos h)

= –2 2h 0

tan hlim

sin h(1 tan h) sin h

1 cos h

= –2 2h 0

sin hlim

sin h(1 cos h sin h) sin h

1 cos h

= –2 h 0

1lim

sin h(cos h sin h) 1

1 cos h

= –2 . 1

(1 0)(0 1) = 2.

Example 32. Let x alim

x xx a

x a

= l , a > 0 and

x alim

x aa x

x a

= m , a > 0 . If l = m then find the value

of 'a'.

Solution l = x alim

x n x x n ae e

x a

= x alim

x n a x( n x n a)e e 1

x( n x n a)

. x( n x n a)

x a

= aa . h 0lim

h(a h) n 1

ah

= aa.

m = x alim

x n a a n xe e

x a

= x alim

a n x x n a a n xe e 1

x n a a n x

. x n a a n x

x a

LIMITS 1.45

= aa . h 0lim

(a h) n a a n (a h)

h

= aa h 0lim a /hh

an a n 1 = aa (ln a 1)

Now l = m a = e2

Example 33. Let

f(x) = 1 1sin (1 {x}).cos (1 {x})

2{x}.(1 {x})

, then find x 0lim f(x)

and x 0lim f(x), where {x} denotes the fractional part of x.

Solution We have f(x)

= 1 1sin (1 {x}).cos (1 {x})

2{x}.(1 {x})

x 0lim f(x) =

h 0lim f(0 + h)

= })h{.(}h{

})h{(cos}).h{(sinlimh

01020101 11

0

= 1 1

h 0

sin (1 h).cos (1 h)lim

2h.(1 h)

= 1

h 0

sin (1 h)lim

(1 h)

.

1

h 0

cos (1 h)lim

2h

(In second limit put cos–1 (1 – h) = 1 – h = cos )

=1

h 0

sin (1 h)lim

(1 h)

.

1

0

cos (cos )lim

2(1 cos )

= 1

h 0

sin (1 h)lim

(1 h)

.

0lim

2sin( / 2)

( > 0)

= sin–1 1 . 1 = /2

andx 0lim f(x) =

h 0lim f(0 – h)

= 1 1

h 0

sin (1 {0 h}).cos (1 {0 h})lim

2{0 h}.(1 {0 h})

= 1 1

h 0

sin (1 h 1).cos (1 h 1)lim

2( h 1).(1 h 1)

= 1

h 0

sin hlim

h

.

1

h 0

cos hlim

2(1 h)

= 1 . / 2

2

=

2 2

.

Example 34. Evaluate )x4(

x2sin1lim

4/x

Solution Let L = )x4(

x2sin1lim

4/x

=

x

4

x2sin1lim

4

14/x

Put x = h4

L =

h

44

h4

2sin1

lim4

10h

= h

))h2(cos1(lim

4

10h

= ))h2(cos1(

))h2(cos1(.

h

))h2(cos1(lim

4

10h

(form 0

0)

= h 0

1 1 cos2hlim

4 h (1 (cos2h))

= ))h2(cos1(h

|sinh|2lim.

4

10h

= 0 0

1 1 | sin h|. 2 lim lim4 h(1+ (cos2h))

h h

= 1 1 1

. 2. P P4 42

where P = h 0

| sin h|limh

.


We show that the limit P does not exist :

L.H.L. = h 0 t 0

| sin h| | sin(0 t |lim limh (0 t)

= t 0

sin tlim 1

t

and R.H.L. = h 0 t 0

sinh | sin(0 t) |lim lim

h (0 t)

= t 0

sin tlim 1

t

L.H.L. R.H.L. Hence P does not exist. The limit L also does not exist.

Special limitsLet us evaluate some limits with the help of thefollowing special limits :

(i) 3x 0

x sin x 1lim

6x

(ii)

x

2x 0

e 1 x 1lim

2x

(iii)x x

3x 0

e e 2x 1lim

3x

(iv) 3x 0

x tan x 1lim

3x


6x 0

e 1 xlim

sin 2x

Solution Let l =3x 3

6x 0

e 1 xlim

sin 2x

= 3x 3

6x 0

e 1 xlim

(2x)

Now put x3 = t

l = t

2t 0

1 e 1 tlim

64 t

again put t = 2y

=

2y

2y 0

1 e 1 2ylim

64 4y

=

y 2 y

2y 0

1 (e 1) 2e 2 2ylim

64 4y

=

2y y

2y 0

1 1 e 1 1 e y 1lim

64 4 y 2 y

l = 1 1 1 .6464 4 2

64.41

2

l l = 128

1.

Example 36. Evaluate x x

x 0

e e 2xlim

x sin x

Solutionx x

x 0

e e 2xlim

x sin x

= x x

x 0 33

e e 2xlim

x sin x. x

x

= 2

1

l

l(say), where

x x

1 3x 0

e e 2xlimx

l and 2 3x 0

x sin xlim

x

l

Now l2 = x 0lim 3x

xsinx = t 0

lim 3t27

t3sint3 Put x =3t

= t 0lim 3

3

t27)tsin4tsin3(t3

= t 0lim

3

3 3t 0

3(t sin t) 4sin tlim

27 t 27t= 2

1 4

9 27l

274

98

2l 61

2l .

Now l1 = x x

3x 0

e e 2xlim

x

Put x = 3y

= y 0lim 3

y3y3

y27y6ee

= y 0lim

( ) ( )- -- + - -3y y y y

3

e e 3 e e 6y

27y

= y 0lim

32y y y

3y 0

8 e 1 1 e e 2ylim

27 2y 9 y

= 191

278

l

278

98 1 l

31

1 l

The required limit = 216.

31

.

LIMITS 1.47


x

1lim x x n 1

x

Solution x

x

1lim x 1 n 1

x

Put x = 1/y

= 2y 0

1 n(1 y)lim

y y

Put ln(1+y) = t 1+y = et

y = et –1 as y0, t0

= t t 2t 0

1 tlim

e 1 (e 1)

= t

2tt 02

e 1 tlim

e 1.t

t

= t

2t 0

e t 1 1lim

2t

(using special limit(ii))

Doing y 0

1 ln(1 y) 1 1 1lim · 0

y y y y y

using

y 0

ln(1 y)lim 1

y

, is not correct.

Example 38. Evaluate 1/x

x 0

(1 x) elim

x

Solution1/x

x 0

(1 x) elim

x

=

ln(1 x) 1x

x 0

e 1 elim

x

, assume M = ln(1 x) 1x

= M

x 0

e 1 e Mlim .M x

=

x 0

e Mlimx

= 2x 0

ln(1 x) xe.limx

=

2e

{using previous example}

Example 39. Evaluate x 0lim 221 x

1)x(sin

1

Solution Solving x 0lim 221

2

2 x1.

)x(sinx.

x1

= x 0lim 22 x

1x1

= 0, is wrong.

The correct way is to put x = sin

x 0lim 221 x

1)x(sin

1

= 2 20

1 1lim

sin

=

0lim

22

22

sinsin

= 0

lim 4

)(sin)(sin

= 30

sin2.lim

= 1 126 3

.

(using special limit(i))


2

2

sinx x {x}.{ x}

xcosx x {x}.{ x}

where {.} denotes the fractional part function.

Solution f(0+) =h 0lim }h}{h{hhcosh

}h}{h{hhsin2

2

= h 0lim )h1(hhhcosh

)h1(hhhsin2

2

= h 0lim hhcosh

hhsin

= h 0lim )hcos1(h

h·h

hsinh 3

3

= 61

·2 = 31

.

Similarly f (0–) = h 0lim h)h1(hhcosh

h)h1(hhsin2

2

= h 0lim )hcos1(h

)hh(sin

= h 0

sin h1

hlim1 cosh

= 2

2 = 1.

Since f (0–) f (0+), the limit does not exist.



2

3

x n 1 x x

x

Solution Let 21 x x = t x 0 , t 1

Also 1 + x2 = t2 + x2 + 2 t x x = 21 t

2 t

.

The given expression =

2

2 3

1 t2 t n t

1 t( )

8 t3

= 2

2 3

2 t n t 1 t

2 t . 1 t( )

8 t3

= 2

3 3

2 t n t 1 t

(1 t) (1 t)

. 4 t2

= 1

2

2

3 3

2 t n t 1 t

(1 t) (1 t)

. 4 t2

Put t = 1 + y; as t 1 , y 0

Limit = 1

2 y 0lim

2

3

2 (1 y) n (1 y) 1 (1 y)

y

Put 1 + y = ez; as y 0, z 0

= 1

2

z 0lim

z 2z

z3e 1

z

2 z e 1 e

. z

= 1

2 z 0

lim

2z z

3

e 2z e 1

z

= 1

2 z 0

lim

z z z

3

e (e e 2z

z

) =

1

2 . l

l = u 0lim

3u 3u

3

e e 6u

27 u

when z = 3 u

= u 0lim

u u 3 u u

3

e e 3 (e e ) 6u

27 u

( )

= u 0lim

32 ue 1

2 u

+ 1

9 l

8

9 l =

8

27 l =

1

3

The required limit = 6

1.

HHHHH1. Evaluate the following limits :

(i)x 0lim

osin xx

(ii)x 0lim x

xtan 3


(i)x 0

sin2xlim

5x (ii) x 0lim sin8xcot3x

(iii) x

sin3xlimsin 2x

(iv) 2x 0

1 cos5xlim

3x


(i)4x

x 0

e 1lim

tan x

(ii) x 1lim

n

m

x 1

x 1

(m, n N)

(iii)x 0lim x

)x1(log2

(iv) 0xlim

ln (1 8x)ln (1 7x)


(i) xx 0

ln(1 x)lim

3 1

(ii) 2x 0

ln sec xlimx


(i)x 0lim

1/m 1/n(1 x) (1 x)

x

(ii)x 0lim

x x

2

a bx 1 x


(i)x 0

sin 2xlim

tan3x

(ii) 2x 0

cosx sec xlim

x

(iii)x 0

cos7x cos9xlim

cosx cos5x

(iv)x 0lim sin 5x

n(1 sin 4x)

e 1

LIMITS 1.49

7. Find whether the following limits exists:

(i)x 0lim xcos1

xcos1x1

(ii)

1

x 0

1cot

xlimx

(iii)x 1lim 1

| x 1|

tan (x 1)

(iv)x 0lim

x4 sin (2 x )


(i)tan x x

x 0

e elim

tan x x

(ii)x 0lim 4 2

n cos x

1 x 1

(iii))xtan(ln

xsinlimx 21

1310

(iv)x 0lim

21 x x 1

sin 4x


(i) 20x x

1xsinx1lim

(ii) x3cos1

x5cos1lim

0x

(iii) )1xln(

xsinx3sinlim

0x

(iv) 20x xtan

)xsinx31ln(lim


(i) xx

xx

0x 56

78lim

(ii))x21(ln

x3sinlim

2

2

0x

(iii))16()13(

)14()15(lim

0

(iv) 0xlim x3tan

1x21


(i)

33

2

0 sintan

)cos1(lim

(ii) 312

2

0 1

11

)x(sinx

)x(xlimx

12. Find 0xlim f(x) if f(x) =

02

01

2

2

xforxx

x

xforx

xcos


(i)x4

xcosxsinlim

4/x

(ii)xln

)1esin(lim

1x

1x

(iii)1x

x3x7lim

23 3

1x

(iv) 3 22/x )xsin1(

xcoslim


(i)x 1

xlim (1 x) tan

2

(ii)

x e

ln x 1lim

x e

(iii) xx 1lim (1 x) log 2

(iv)3 2

x 1

x x 2lim

sin(x 1)

GGGGG15. Show that

x

)x1sin()x1sin(lim

0x

= 2cos 1.

16. Prove that 3x 0

1 tan x 1 sin xlim

x

= 4

1

17. Evaluate x x

3x 0

cos(xe ) cos(xe )lim

x

18. Evaluate xsin

x])tan([)x]tan([lim

2

2222

0x

where [.] denotes the greatest integer function.


19. Evaluate m

n

0 )(sin

)sin(lim

(m and n positive integers)

20. Evaluate 20h h

asin)hasin(2)h2asin(lim


(i)x 0lim

n(1 a sin x)

sin x

(ii)x 0lim

4

1 cosx

sin 3 x

(iii) x 0lim

2

2

sin3x

ln cos(2x x)

(iv) 3x 0

2sin x sin 2xlimx


(i)x 0lim

2 3

2 4

3sin x x x

tan x 2sin x 5x

(ii) 521

31

0x xxxtan3

xsinxsin2xcos1lim

(iii)x 0lim

1 2 1 2sin 2x (sin x) (tan x)

3x


(i) 3

3

1 2x 0 5 x

sin x ln(1 3x)lim(tan x ) e 1

(ii)2x 3x

2x 0

e 3limxsin sin x2

(iii) x 0lim

sec x/2

sec x

log cosx

log cosx / 2

24. Evaluate .2,1n,)–x(5

)x–(7cos1lim

nx


(i)xcos21

xtan1lim

2

3

4/3x

(ii)

x

x 1

x2 xlim

1 cosx


(i) 22

22

sinsinlim

(ii)y a

y a ylim sin . tan2 2a

27. Evaluate the one-sided limits and find whetherthe limit exists:

(i)x 1lim

xsin(x [x])

x 1

(ii)x 1lim

1 cos2(x 1)

x 1

28. Evaluate 22x )2x(

)2xsin(}x{lim

, (where {.} denotes

the fractional part function.)29. Evaluate the following limits :

(i) 1

x 0

cos (1 x)lim

x

(ii)

442

21

xsin

xcos

xcos

xsin

limx

(iii) x a

x a

x alim (a 0)

x a


(i))xx4x31ln(

)x2x3x1ln(lim

32

32

1x

(ii)xlnx

1xlim

x

1x


(i)1x

xcoslim1

1x

(ii) 4x 1

ln xlim

x 1 32. Evaluate the following limits :

(i)x /4lim

54 2 (cosx sin x)

1 sin 2x

(ii) x /6lim

sin(x / 6)

3 2 cosx


(i)2 3

102 3x 0

10

log (1 2x 3x 4x )limlog (1 x 2x 7x )

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