limits · a-479 indra vihar, kota rajasthan 324005 ... in the notion of limits note very carefully...

Topic Page No.

Theory 01 - 04

Exercise - 1 05 - 12




Answer Key 22 - 23

Contents

LIMITS

SyllabusStandard Limits, Indeterminant forms, L-H Rule, Series Expansion

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Page No. # 1Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005

LIMIT OF FUNCTION

KEY CONCEPTS1. Definition :

The notion of limit is what separates calculus from other branches of mathematics.

It is very important to note that limits of a function at a point ‘x = a’, exists only when

the function is defined in the immediate neighbour- hood of ‘a’.

For eg. xsinLim 12x

-

® is non-existent, & 6/xsinLim 1

2/1xp=-

®.

Also it is necessary to note that if f(x) is defined only on one side of ‘x = a’, one sided limitscan be evaluated & if f(x) is defined on either side of ‘a’both sided limits are to be considered.

As in 0xcosLim 1

1x=-

® -, though f(x) is not defined for x>1, even in it’s immediate vicinity..

The most important aspect to be considered is that ax

Lim®

necessarily implies

‘x ¹ a’In the notion of limits note very carefully that we are not at all concerned with what happens to thefunction at x = a

THE THEORETICAL DEFINITION OF LIMITS

l=®

)x(flimax

, if $>ÎÚ ,0 some d > 0 such that .|ax|0|)x(f| d<-<ÚÎ<- l

2. Evaluation of Left Hand Limits (LHL) and Right Hand Limits (RHL) :When 'x' approaches to 'a' i.e. when x ® a then this variable 'x' on R–line can do this job in two ways oneform the left hand side i.e. from negative side of the fixed point ‘a’ and second form the right hand side (i.e.from positive side) of ‘a’ as shown in the figure below

a(Left hand side) (Right hand side)x x

R-line-¥ +¥

LHL is denoted by )x(flim0ax -®

or by x alim f(x)

-® and similarly RHL by )x(flim

0ax +® or by )x(flim

ax +® and are

determined as (i.e., working rule is)

LHL = x alim f(x)

-® = ( )

h 0lim f a - h

® and )ha(flim)x(flimRHL

0hax+==

®® + ; where h > 0

Existence of Limit :The limit of a function at a point exists if both left and right limits of the function at that point exist and are

equal. Thus ( ) ( ) ( )Û® ® ®- +lim f x exists lim f x = lim f xx a x a x a

Note :– (i) Limit of a function at a point (if exists) is unique.(ii) If at some point of a continuous function its value and limit both exist, then they are necessarily equal.

3. Indeterminate Forms :The concept of limit was evolved to deal with indeterminate forms of some functions at some specific points.

For example the function f(x) = 2x4x2

--

is ready to give its values at x = 0 , –1, 1, 3, 2 ..... etc.


Q f(0) = 2, f(–1) = 1, f(1) = 3, f ( 2 ) = (2 + 2 ).......etc.} but it fails to give its value at x = 2, as

f (2) = 00

22422

=--

(known as an indeterminate form)

(Note, that 00

is capable of assuming any value, that is why it is called ‘indeterminate’), So in this case

we find the limit, as follows.

2x4xlim)x(flim

2

2x2x --

=®®

= x 2

(x 2) (x 2)limx 2®

- +-

= x 2lim

® (x + 2) = 4 ....(1)

Q f (x) = 2x

)2x()2x(2x4x2

-+-

=--

= x + 2 Þ f(2) = 4 ....(2) is a wrong calculation,

will be explained in the class room. Following indeterminate forms are in our course, namely (i) 00 (ii)

¥¥

(iii) 0´ ¥ (iv) ¥-¥ (v) ¥1 (vi) 0¥ (vii) 00

4. Properties Of Limits :

· Let f and g be two real functions with domain D. We define four new functions ff g ,fg ,g

± , g ¹ 0 on domain

(D) by setting

· ( )( ) ( ) ( ) ( )( ) ( ) ( )f g x f x g x f g x f x g x± = ± · =

· ( ) ( )( )

æ ö=ç ÷

è ø

f xf x ,g g x if ( ) ¹g x 0 for any Îx D .

Following are some results concerning the limits of these functions.

Let ( )®

lx alimf x = and ( )

®x alimg x = m . If l and m exist.

(i) Sum and Difference Rule ( )( ) ( ) ( )x a x a x alim f g x lim f x lim g x m

® ® ®± = ± = ±l

(ii) Product Rule ( )( ) ( ) ( )x a x a x alim fg x lim f x lim g x m

® ® ®= = l

(iii) Quotient Rule ( )( )( )

x a

x ax a

lim f xflim xg lim g x m

®

®®

æ ö= =ç ÷

è ø

l

, provided ¹m 0

(iv) Constant Multiple Rule ( ) ( )® ®

=x a x alimKf x K.lim f x , where K is constant

(v) Modules Rule ( ) ( )x a x alim f x lim f x

® ®= = l

(vi) Power Rule ( )( ) ( )g x

x alim f x

®=

)x(glim

axax)}x(flim{ ®

®

(vii) Composite Function Rule ( ) ( )( ) ( )® ®

= =x a x alim fog x f limg x f m In particular

(a) ( ) ( )( )x a x alimlogf x log lim f x log

® ®= = l

(b) ( ) ( )x alim f xf x

x alime e e®

®= = l

(c) If ( )®

= +¥x alim f x or -¥ , then

( )®=

x a

1lim 0f x


5. Some Standard Limits :

(i) If ( ) ®f x 0 , when ®x a , then :

(a)( )

( )®=

x a

sin f xlim 1

f x (b) ( )®

=x alimcos f x 1

(c)( )

( )®=

x a

tanf xlim 1

f x (d)( )

( )®

-=

f x

x a

e 1lim 1f x

(e)( )

( ) ( )®

-= >

f x

x a

b 1lim lnb, b 0f x (f)

( )( )x a

ln 1 f xlim 1

f x®

é ù+ë û =

(g) ( ) ( )1/ f x

x alim 1 f x e

®é ù+ =ë û

(h) ( )®

= >x alim f x A 0 and ( )

®f =

x alim x B then ( ) ( x) B

x alim f x A

f

®é ù =ë û

(ii) So from above Results :

(a)®

=x 0

sinxlim 1x

(b) x 0lim cosx 1®

=

(c)®

=x 0

tanxlim 1x

(d)®

-=

x

x 0

e 1lim 1x

(e) ( )x

x 0

a 1lim lna, a 0x®

-= > (f) ( )

®

+=

x 0

ln 1 xlim 1

x

(g) ( )®

+ =1/ x

x 0lim 1 x e (h)

®¥

æ ö+ =ç ÷è ø

x

x

1lim 1 ex

(i) -

®

-=

-

m mm 1

x a

x alim max a (j) -

®

-=

-

m mm n

n nx a

x a mlim anx a

6. Methods of Evaluation of Limits :

(i) Substitution method (ii) Factorization method

(iii) Rationalization or double rationalization (iv) Expansion Method

(v) When x ® ¥ (vi) Simplification

(vii) Sandwich Theorem

(viii) Evaluation of a limits of the form 0 × ¥ , ¥ – ¥ , 0° , ¥ °

(ix) Evaluation of a limits of the form ¥1

(i) Substitution method : In some cases limit of a function can be found by simple substitution if on substitutionthe function does not take indeterminate form.

(ii) Factorization method : If f(x) and g(x) are polynomials and ( )g a 0¹ , then we have

( )( )

( )( )

( )( )

x a

x ax a

lim f xf x f alim

g x lim g x g a®

®®

= =

(iii) Rationalization or double rationalization :

In this method we rationalize the expression with the help of factor inside the square root and put the valueof x.


(iv) Expansion Method : Using expansion of functions :

If ®x 0 and there is atleast one function in the given expression which can be expanded, then we expressNr and Dr in the ascending powers of x and remove the common factor there.The following expansions of some standard functions should be remembered :

(a) = + + + +2 3

x x xe 1 x ......2 3 (b) - = - + - +

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

2 3

(c) ( ) ( ) ( )= + + + +

2 3x xloga xloga

a 1 xloga ......2 3 (d) ( )+ = - + -

2 3x xlog 1 x x ......2 3

(e) ( )- = - - - -2 3x xlog 1 x x .....

2 3 (f) = - + -3 5x xsinx x ......3 5

(g) = - + - +2 4 6x x xcos x 1 ......2 4 6 (h) = + + +

35x 2tanx x x ......

3 15

(i) = + + +3 3x xsinhx x ......3 5 (j) = + + +

2 4x xcoshx 1 ......2 4

(k) = - + -3

5xtanhx x 2x .......3 (l)

- = + + +3 5

1 x 9xsin x x ......3 5

(m)- æ öp

= - + + +ç ÷è ø

3 51 x 9xcos x x .......

2 3 5 (n) - = - + - +3 5 7

1 x x xtan x x ......3 5 7

(o) ( ) ( )-+ = + + +n 2n n 1

1 x 1 nx x ......2 (p) ( )1/ x 2x 111 x e 1 x .....

2 24æ ö+ = - + +ç ÷è ø

(v) When x ® ¥This type of problems are solved by the taking the highest power of the terms tending to infinity as commonin numerator and denominator then take the limit of the function.

(vi) Simplification:In this method the indeterminate form is removed by simplifying the expression.

(vii) Sandwich Theorem : Suppose that ( ) ( ) ( )g x f x h x£ £

for all x in some open interval containing c, except possiblyat c itself. Suppose also that

( ) ( )x c x clim g x limh x L

® ®= = ; Then ( )

x clim f x L

®=

This is Sandwich Theorem or Squeeze Play Theorem.(viii) Evaluation of a limits of the form 0 × ¥ , ¥ – ¥, 0° & ¥°.

(ix) Evaluation of a limit of the form ¥1 :

(i) If ( ) ( )® ®

= =x a x alim f x lim g x 0 , then ( ) ( )

( )( )®

é ùê úê úë û

®é ù+ =ë û

x a

f x1 lim

g xg xx alim 1 f x e

(ii) If ( ) ( )® ®

= = ¥x a x alim f x 1,lim g x , then ( ) ( ) ( )( ) ( ) ( ) ( )

®é ù-ë û

® ®é ùé ù = + - =ë û ë û

x ag x lim f x 1 g xg x

x a x alim f x lim 1 f x 1 e

Particular case:

(i) ( ) l

®+ l =1/ x

x 0lim 1 x e (ii) l

®¥

læ ö+ =ç ÷è ø

x

xlim 1 e

x


PART - I : OBJECTIVE QUESTIONS

* Marked Questions are having more than one correct option.

Section (A) :

1. lim | |x

xx® 0 is equal to

(A) 0 (B) 1 (C) – 1 (D) does not exist

2.1x

im®

l (1 - x + [x - 1] + [ 1 - x]) is equal to (where [.] denotes greatest integer function)

(A) 0 (B) 1 (C) - 1 (D) does not exist

3. 1im

x®-l

2 2 31

x xx- -

+(A) 0 (B) – 4 (C) ¥ (D) does not exist

4. 1xim®

l

1x

100x100

1k

k

-

-÷÷

ø

ö

çç

è

æå

= is equal to

(A) 0 (B) 5050 (C) 4550 (D) - 5050

5. Limitx ® 9

34 2 2

-

- -

xx

is

(A) 32

(B) 31

(C) – 31

(D) none

6. 1xLimit

® 1x22x5

--

-- =

(A) 2 (B) 1/2 (C) - 1 (D) none

7. Limitx ® ¥ ( )( ) ( )x a x b x+ + - =

(A) ab (B) a b+

2(C) ab (D) none

8. axLimit

® )ax()2a()2x( 3/53/5

-+-+

=

(A) (a + 2)2/3 (B) 35

(a + 2)3/2 (C) 35

(a + 2)2/3 (D) 53

(a + 2)3/2


9.¥®n

iml n

n

)1(n4)1(n3

---+-

is equal to (n Î N)

(A) – 43

(B) – 43

if n is even ; 43

if n is odd

(C) not exist if n is even ; – 43

if n is odd (D) 1 if n is even ; does not exist if n is odd

10. ¥®ximl

1xx

1xx1sinx

2

3

++

++ is equal to

(A) 0 (B) 21

(C) 1 (D) none of these

11. Limitx ® 0

37 5

x xx x

+-

=

(A) 2 (B) 1/6 (C) 0 (D) does not exist

12. ¥®niml 5 3 2

5 2 3

1 2

2 3

n n n

n n n

+

+

+ -+ +

is equal to

(A) 5 (B) 3 (C) 1 (D) zero

13. 5xLimit

® ]x[x20x9x2

-+-

where [x] is the greatest integer not greater than x:

(A) 1 (B) 0 (C) 4 (D) none

14. lim tansinxx x

x x®-

-02

3 is equal to

(A) –1/3 (B) 1/2 (C) 1 (D) 0

15. limx®¥

2cossin

x xx x-+

=

(A) 0 (B) 1 (C) ¥ (D) none

16. =-

-++++® 1x

5)xxxxx(Lt5/14/13/12/1

1x

(A) 60138

(B) 60139

(C) 60140

(D) 60137

17. 0xim®

l

xsinxxcos–e

3

2

2x–

is equal to

0xim®

l

xsinxxcos–e

3

2

2x–

(A) 41

(B) 61

(C) 121

(D) 81


18. ¥®ximl

xx

x+-

æèç

öø÷

+22

1

is equal to

(A) e4 (B) e -4 (C) e2 (D) none of these

19. ¥®ximl

x

2

2

2x4x1x2x

÷÷ø

öççè

æ

+-+-

is equal to

(A) 1 (B) 2 (C) e2 (D) e

20. =-® 20x x

BxcosAxcosLt

(A) 2BA 22 + (B) 2

BA 22 - (C) 2AB 22 - (D) none of these

21. +® 0ximl ( ) x

52 xtan1+ is equal to

(A) e5 (B) e2 (C) e (D) none of these

22. The limiting value of xsin1

)x(cos as x ® 0 is:(A) 1 (B) e (C) 0 (D) none of these

23.sinlim

®¥x

xx

(A) 0 (B) 2 (C) does not exist (D) 1

24.3x

1

0x xtanxLim ÷

ø

öçè

æ+®

is equal to –

(A) e1/3 (B) e–1/3 (C) Does not exists (D) 0

25. 2xim®

l( )

)1x(n1esin 2x

---

l is equal to

(A) 0 (B) - 1 (C) 2 (D) 1

26. 3xim®

l

)9x(

)2x(n)27x(2

3

-

-+ l

is equal to

(A) - 8 (B) 8 (C) 9 (D) - 9

27. If úûù

êëé -+=

¥®xx2xlim 2

xl & { }xx2xlimm 2

x+-=

-¥® , where [ ] & { } represent integral and fractional

part respectively then m+l is equal to–

(A) 0 (B) 1 (C) 2 (D) 3

28.xsinxtanxcos21x

)xcos1(n)x1(nxsinxcos1Lim 3422

333

0x +++-+++++-

®

ll

(A) 43

(B) ln2 (C) 42nl

(D) 3/2


29. The value of 0xim®

l ( )

)xsin1(n)x1(nsin

++

l

l is equal to

(A) 0 (B) 21

(C) 41

(D) 1

30. 0xim®

l

÷÷ø

öççè

æ+÷÷

ø

öççè

æ-

3x1n

pxsin

)14(2

3x

l

is equal to

(A) 9 p (ln 4) (B) 3 p (ln 4)3 (C) 12 p (ln 4)3 (D) 27 p (ln 4)2

31.2

xim

p®

l úúúú

û

ù

êêêê

ë

é p-

xcos2

x

is equal to (where [ . ] represents greatest integer function)

(A) – 1 (B) 0 (C) – 2 (D) does not exist

32.¥®n

iml n cos p4 n

æèç

öø÷ sin p

4 næèç

öø÷ is equal to:

(A) 3p

(B) 4p

(C) 6p

(D) none of these

33. If 2 3x

x-

< f(x) < 2 52

2x x

x+

then Limitx ® ¥ f(x) :

(A) is equal to 1 (B) is equal to 2 (C) is equal to zero (D) does not exist

34. -®axlim | |x

axa

3 3- é

ëêùûú

æ

èçç

ö

ø÷÷ (a > 0), where [x] denotes the greatest integer less than or equal to x is

(A) a2 + 1 (B) a2 - 1 (C) a2 (D) – a2

PART - II : MISCELLANEOUS OBJECTIVE QUESTIONS

1. Evaluate the following limits :

(i) 2xim®

l (x + sin x) (ii) 3xim®

l (tan x – 2x)

(iii)43x

im®

l x cos x (iv) 5xim®

l xx


(i) x4x

1x5Lim 24x +-+

-®(ii) ú

ú

û

ù

êê

ë

é÷÷ø

öççè

æ

--

-

®

1

3

3

2x 8xx4xLim (iii) h

xhxLim0h

-+®

(iv) úû

ùêë

é

+--

-® 2x3x1

)2x(x1Lim 222x

(v) xcos1xcos1Lim

2

0x --

®

(vi) 330x x1x1x1x1Lim

--+--+

®(vii)

Lim xxx®

--p

4

11 2

tansin


3. (i) If f(x) = îíì

³-<+

1x,3x21x,1x

, evaluate 1xim®

l f(x).

(ii) Let f(x) = îíì

³-<l+

1x,3x21x,x

, if 1xim®

l f(x) exist, then find value of l.

4. Which of the followings are indeterminate forms. Also state the type.

(i) +®0ximl

x]x[

, where [ . ] denotes the greatest integer function

(ii)-¥®x

iml 1x2 + – x (iii)2

xim

p®

l (tan x)tan2x

(iv) +®1ximl { }( ) nx

1

x l , where { . } denotes the fractional part function

5. Evaluate the following limits, if exists

(i) 0xim®

l x5cos1x4cos1

--

(ii)6

xim

p®

l

6x

xcosxsin3p-

-

(iii) 0xim®

l xsinx3x2x3tan

2--

(iv) 0xim®

l x

asina)xasin()xa( 22 -++

6. Evaluate each of the following limits, if exists

(i)1x

1x3xim3

1x -+-

-®l (ii)

4x2x5x5x2xx4im 36

23

1x --+-+-

®l

(iii) axim®

l x2xa3x3x2a

-+

-+, a ¹ 0

7. Evaluate ( ) ÷÷ø

öççè

æ-++++

¥®x)4x()3x()2x()1x(im 4

1

xl

8. If 0xim®

l 3

x

xcexcosxsinba +-+

exists, find the values of a, b, c. Also find the limit


(i)4

xim

p®

l (tan x)tan2x (ii) ¥®x

iml x

x31x21

÷ø

öçè

æ++

(iii) 1xim®

l ( ) 2xsec

nx1p

+ l (iv) +®0ximl ( ) 2xx

(v)-p

®2

x

iml (tan x)cosx

(vi) -®1ximl ([x])1–x , where [ . ] denotes greatest integer function


10. If 1xim®

l ( ) 1xc

2bxax1 -++ = e3 , then find conditions on a, b and c.

ASSERTION/REASON TYPE

11. STATEMENT-1 : 0xim®

l x1

x4

tan ÷÷ø

öççè

æ÷øö

çèæ +

p = e

STATEMENT-2 : axim®

l (1 + f(x))g(x) = )x(g.)x(fimaxe ®

l , if axim®

l f(x) = 0 and axim®

l g(x) = ¥.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

(C) Statement-1 is True, Statement-2 is False.

(D) Statement-1 is False, Statement-2 is True.

12. STATEMENT-1 : 32

x3x2x3x7x3x2im 24

34

x=

++

++¥®

l .

STATEMENT-2 : If P(x) and Q(x) are two polynomials with rational coefficients, then

)x(Qinxofpowerhighestoftcoefficien)x(Pinxofpowerhighestoftcoefficien

)x(Q)x(Pim

x=

¥®l

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

(C) Statement-1 is True, Statement-2 is False.

(D) Statement-1 is False, Statement-2 is True.

COMPREHENSIONS :

Read the following passage carefully and answer the questions.

Comprehension # 1

Consider two functions f(x) = ¥®niml

n

nxcos ÷÷

ø

öççè

æ and g(x) = – b4x , where b = ¥®x

iml ÷øöç

èæ +-++ 1x1xx 22 ,

then

13. f(x) is

(A) 2xe- (B) 2

x2

e-

(C) 2xe (D) 2

x2

e

14. g(x) is(A) – x2 (B) x2 (C) x4 (D) –x4


Comprehension # 2

Let f(x) = p – cot–1(tanx) ; x Î -FHGIKJ

p p2 2

, and g(x) = –tan–1(cotx) ; x Î (0, p). Now consider a circle x2+

y2 – py = 0. y = f(x) cuts the circle at points C and D and y = x cuts the circle at points A and B.

15. limx 0®

-sin{ ( ) ( )}f x g xx

2 2 is-

(A) p (B) p2

(C) 2p (D) doesn't esists

16. Area of quadrilateral ABCD is-

(A) p2

4 2(B) ( )2 1

8

2+

p(C) ( )2 1

8

2-

p(D)

p2

8

17. Let l = limx®-

-+ +p p2

1f x

f x( )

( ) , x

sin1 cos(f(x))limcos1 sin(g(x))®

++l

is-

(A) cot1 (B) 1 (C) –cot1 (D) none of these

Comprehension # 3

AB is arc of a circle, C is its mid-point, O is the centre of this circle. Tangents at A and B meet in T. Tangent at

C meets tangents at A and B in D and E respectively. Further Ð =AOC q and Radius of the circle is R

18. Area of triangle ABT, if qp

=3

is-

(A) 32

2R (B) 2 3 R2 (C) 3R2 (D) 3 34

2R

19. Area of triangle TDE is-

(A) 2 12

2 2R ( cos )sin

- qq

(B) R2 2

21( cos )

sin

- q

q

(C) R2 212

( cos )sin- q

q(D) R2 21( cos )

sin- q

q

20. lim0q®

Area of ABTArea of TDE

( )( )DD is equal to-

(A) 2 (B) 12

(C) 6 (D) 4


Match the Column :

This section contains 3 questions numbered (26-28). Each question contains statements given in twocolumn which have to be matched. Statements (A,B,C,D) in Column–I have to be matched withstatement (P,Q,R,S) in Column–II.

21. Match the followingColumn - I Column - II

(A)

24

3 3 3lim1 2 .....n

nn®¥

æ öç ÷+ + +è ø

is (p) 64

(B) limx®

--1

32 11

xx

is equal to (q) 29

(C) limx 2®

+ --

x xx

3 2 122

equals (r) 12

(D) limx ® 0e x

x

xsin - -12 is equal to (s) 16

22. Match the Column :A right angled triangle has legs 1 and x. The hypotenuse is y and the angle opposite to the side x is q.Column - I Column - II

(A) 2

Lim y x-p

q®

-(p) 0

(B) 2

Lim y x-p

q®

-(q) 1/2

(C) 2 2

2

Lim y x-pq®

-(r) 1

(D) 3 3

2

Lim y x-pq®

-(s) ¥

23. Match the followingColumn-I Column-II

(A) limx 0®

-1 22

cos xx equals (p) an even number

(B) limx 0® -

xx x

3

sin equals (q) an odd number

(C) limn®¥

=

=

å

å

n r

r

r

n

r

n

2

1

3

1

equals (r) a prime number

(D) limx®

+ + + --1

2 3 4 8 41

x x x xx

equals (s) a composite number


PART - I : OBJECTIVE QUESTIONS

1. Which of the following limit is not unity,

(A) 2

2

0x x)xsec(sinLim

pp

®

(B) })x{}x({Lim1x

-+® , where { } denotes the fractional part function

(C) ( )2sin.2)1(

2 ---

+® xxxnLim

x

l(D) t

)t(tansinLim0t®

2. ¥®ximl ÷÷

ø

öççè

æ÷øö

çèæ +-

x11nxx 2

l is equal to :

(A) 21

(B) 23

(C) 31

(D) 1

3. limx 0®

-( ) tane xx

x 1 2

3

(A) does not exist (B) exists and equals 0(C) exists and equals 2/3 (D) exists and equals 1

4. A weight hangs by a spring and is caused to vibrate by a sinusodial force. Its displacement f(t) attime t is given by an equation of the form-

f tc k

kt ct( ) [sin( ) sin( )]=-

-a

2 2 Then limc k®

f t( ) is-

(A) -a t

kk t

22cos( ) (B) -

a tk

kt2

cos( ) (C) at

kkt

2sin( ) (D)

atk

kt2

2cos( )

5. The lim it of 1 2 44 1n

k k kk

n( )( )+ +

=å as n ® ¥

(A) exists and equals 1/4 (B) exists and equals 0(C) exists and equals 1/8 (D) does not exists

6. +® 0ximl

x)x1(cos 1 --

is equal to

(A) 21

(B) 2 (C) 1 (D) 0


7. The value of )x(tann

1

4x

])x[1(im l

l +p

® is equal to (where [ . ] denotes the greatest integer function)

(A) 0 (B) 1 (C) e (D) e-1

8. l=--

® xsinx)xcos1(xlim

0x and m=÷

øöç

èæ

¥® x1tanxlim

x then–

(A) l & m both are prime (B) l is prime and m is composite

(C) l is composite & m is prime (D) l is prime and m is neither prime nor composite

9. 5

33

0x xxsinxtanLim -

® is equal to

(A) 3/2 (B) –3/2 (C) 0 (D) Does not exists

10. n

x

0x x1xsinxtaneLim

3-+-

® exists and is non-zero, then the value of n is

(A) 1 (B) 3 (C) 2 (D) 0

11. If k is an integer such that n n

n

k klim cos cos 04 6®¥

é ùp pæ ö æ ö- =ê úç ÷ ç ÷è ø è øê úë û

then

(A) k is divisible neither by 4 nor by 6(B) k must be divisible by 12 but not necessarily by 24(C) k must be divisible by 24(D) either k is divisible by 24 or k is divisible neither by 4 nor by 6

12. limx

x

x

e

e

n

n®

-

+0

1

11

1

exists, n NÎa f , then ‘n’ can be

(A) 1 (B) 2 (C) 3 (D) 7

13. limcot

xx

x®

- FH IK0

1 1

(A) Exists and is equal to one(B) does not exists as R.H.L is 1 and L.H.L is -1(C) does not exists as R.H.L and L.H.L both are non–existent.(D) does not exists as R.H.L exists but L.H.L does not.

14.úûù

êëé

úûù

êëé-úû

ùêëé

®x

xtanx

xsinxsin

x

lim0x is equal to,( [ ] represents the greatest integer function)

(A) 1 (B) -1 (C) 0 (D) Does not exists


15. im0x

l

®( )

úúû

ù

êêë

é-

xxsine1 x

, where [×] represents greatest integer function, is equal to

(A) – 1 (B) 1 (C) 0 (D) does not exist

16. Let a, b be the roots of equation ax2 + bx + c = 0, where 1 < a < b and imxx

0

l® cbxax

cbxax2

2

++

++ = 1, then which of

the following statements is incorrect(A) a > 0 and x0 < 1 (B) a > 0 and x0 > b(C) a < 0 and a < x0 < b (D) a < 0 and x0 < 1

17. If im0x

l

® )xsinbx(xax3

-+ = 1, then constants ‘a’ and ‘b’ are (where a > 0)

(A) b = 1, a = 36 (B) a = 1, b = 6 (C) a = 1, b = 36 (D) b = 1, a = 6

18. If l = ¥®ximl (sin 1x + – sin x ) and m = ¥-®x

iml [sin 1x + – sin x ], where [.] denotes the greatest integer

function, then :(A) l = m = 0 (B) l = 0 ; m is undefined(C) l, m both do not exist (D) l = 0, m ¹ 0 (although m exist)

19. 3333

2222

n n.....3211.n.....)2n(3)1n(2n1Lim

++++++-+-+

¥® is equal to :

(A) 31

(B) 32

(C) 21

(D) 61

20. +®0ximl xsinlog

2xsin

is equal to

(A) 1 (B) 0 (C) 4 (D) 41

MULTIPLE OPTIONS CORRECT

21. Let f(x) = ïî

ïíì

<£

<£+

2x1,ax

1x0,ax21 , if 1x

im®

l f(x) exists, then value of a is :

(A) 1 (B) – 1 (C) 2 (D) – 2

22. Let f(x) = ]x[x20x9x2

-+-

(where [x] denotes greatest integer less than or equal to x), then

(A) -®5ximl f(x) = 0 (B) +®5x

iml f(x) = 1

(C) 5xim®

l f(x) does not exist (D) none of these

23. If 0xim®

l ( )x2

2bxax1 ++ = e3, then possible values of a and b is/are :

(A) a = 3, b = 0 (B) a = 23

, b = 21

(C) a = 23

, b = 23

(D) a = 23

, b = 0


24. If 0x

im®

l ( ) x1

bxsinaxcos + = e2 , then the possible values of ' a ' & ' b ' are :(A) a = 1 , b = 2 (B) a = 2 , b = 1 (C) a = 3, b = 2/3 (D) a = 2/3 , b = 3

25. If f(x) = |x|xx2cos2cos

2 -

-, then

(A) 1xim

-®l f(x) = 2 sin 2 (B) 1x

im®

l f(x) = 2 sin 2

(C) 1xim

-®l f(x) = 2 cos 2 (D) 1x

im®

l f(x) = 2 cos 2

26. If l = 0xim®

l3x

xsinb)xcosa1(x -+ = 0x

im®

l2x

xcosa1+ – 0x

im®

l 3xxsinb

, where l Î R, then

(A) (a, b) = (–1, 0) (B) a & b are any real numbers

(C) l = 0 (D) l = 21

27. If f(x) = 6x32x2

-+

, then

(A) -¥®x

iml f(x) = – 31 (B)

¥®ximl f(x) =

31 (C)

-¥®ximl f(x) =

31 (D)

¥®ximl f(x) = –

31

28. If f(x) = xsin|x| , then

(A) -®0ximl f(x) = 1 (B) +®0x

iml f(x) = 1

(C) 0xim®

l f(x) = 1 (D) limit does not exist

29.Ax)1ax(im n

n

x +

+¥®

l is equal to

(A) an if n Î N (B) ¥ if n Î Z– & a = A = 0

(C) A1

1+

if n = 0 (D) an if n Î Z–, A = 0 & a ¹ 0

30. If 0xim®

l 3xxsinax2sin +

= p (finite), then

(A) a = – 2 (B) a = – 1 (C) p = – 2 (D) p = – 1

PART - II : SUBJECTIVE QUESTIONS

1. Evaluate the following limits.

(i) limx

x xx x x®

- +- + +1

4 3

3 2

3 25 3 1

(ii) lim log logx

x x x xx®

- + --1

2 11

(iii) ])x[x(lim0x

-®

(where [ .] is G..I.F.)

(iv) 3x |x |lim

7x 5|x|x 0+

-®(v) lim ( )

xx

x x®-

++

-RST

UVW32 9 1

31

3 (iv) lim

/x e x® -01

1 1

(v) lim| |x

xx x® +0

32 2

(vi) limx

xx x®

-+ -2

4

2

43 2 8

(vii) lim/

/xee

x

x®+-0

1

1

11



(i) ®x 0Lim

2 4

x

ln(1+ x + x )(e -1)x (ii) Evaluate

® +x 0Lim In(1+ x) - x

x

3. Evaluate: Lim n xx

Lim n xxx x® ®

LNM

OQP + L

NMOQP0 0

sin tan where [*] denotes the greater function and n IÎ - 0l q

4. Evaluate : ÷øö

çèæ

- -® xx20x eexsin2.

x)x3(cosnlim l

5. xsinxxsin

0x xxsinlim -

®÷øö

çèæ

6. Evaluate.Lim a a a anx

x x xn

x nx

®¥

+ + +FHG

IKJ

11

21

31 1/ / / /......

where a a a an1 2 3 0, , .... >

7. )0a(aaaalim xx

xx

x>

+

--

-

¥±®

8. Lim x x x xx®¥

+ + -LNM

OQP =

9. 30h hasin)hasin(3)h2asin(3)h3asin(lim -+++-+

®

10. Evaluate:®0

limn

+ - + - + ++ + + +2 2 2 2

1. 2( 1) 3( 2) ...... .11 2 3 ......

n n n nn

11. 40x x)xcos1cos(1lim --

®

12. 3x 0

x n(1 x) 2cosx 2Limx®

+ + -l is equal to :

13. 2x 0

1 cos x cos2x cos3x cos4xLimx®

- is equal to :

14. xcosxxsinxlim

x ++

¥®

15. ( )xcos1xcoslimx

-+¥®


PART-I IIT-JEE (PREVIOUS YEARS PROBLEMS)

1. 1xim®

l 1x

)1x(2cos1-

-- [IIT – 1998]

(A) exists and it equals 2

(B) exists and it equals - 2

(C) does not exist because x - 1 ® 0

(D) does not exist because left hand limit is not equal to right hand limit.

2. 0xim®

l2)x2cos1(

xtanx2x2tanx-

- is equal to [IIT-JEE-1999]

(A) 2 (B) - 2 (C) 21

(D) - 21

3. For x Î R, ¥®ximl

x

2x3x

÷÷ø

öççè

æ+-

equals to [IIT-JEE - 2000]

(A) e (B) e -1 (C) e-5 (D) e5

4. 0xim®

l2

2

x)xcos(sin p

equals to [IIT-JEE-2001]

(A) - p (B) p (C) 2p

(D) 1

5. The integer ' n ' for which 0xim®

l n

x

x)ex(cos)1x(cos --

is a finite non-zero number, is [IIT-JEE-2002]

(A) 1 (B) 2 (C) 3 (D) 4

6. If 0xim®

l ( )

2xxnsinxtanxn)na( --

= 0, where n is a non-zero real number, then a is equal to [IIT-JEE-2003]

(A) 0 (B) n

1n+(C) n (D) n +

1n

7. Let L = 4

222

0x x4xxaa

im---

®l , a > 0. If L is finite, then [IIT-JEE-2009]

(A) a = 2 (B) a = 1 (C) L = 641

(D) L = 321


SUBJECTIVE

8. Find the values of constants a, b & c so that 2xsinx

cxe)x1(nbaxeim2

xx

0x=

++- -

®

ll . [IIT – 1997, 6]

PART-II AIEEE (PREVIOUS YEARS PROBLEMS)

1. 0xim®

l x2

x2cos1- is equal to [AIEEE– 2002]

(A) 1 (B) – 1 (C) 0 (D) does not exist

2. ¥®ximl

x

2x3x

÷ø

öçè

æ+-

, x Î R, is equal to [AIEEE– 2002]

(A) e (B) e–1 (C) e–5 (D) e5

3. Let f(2) = 4 and f¢(2) = 4, then 2xim®

l 2x

)x(f2)2(fx--

is equal to

(A) 2 (B) – 2 (C) – 4 (D) 3

4. If f(1) = 1, f¢(1) = 2, then 1xim®

l 1x1)x(f

-

- is equal to

(A) 2 (B) 4 (C) 1 (D) 21

5. ¥®ximl ]x[

]x[nxn -l, n Î N, where [x] denotes greatest integer less than or equal to x. [AIEEE– 2002]

(1) has value – 1 (2) has value 0 (3) has value 1 (4) does not exist

6. ¥®ximl

x

2

2

3xx3x5x

÷÷ø

öççè

æ

++++

is equal to [AIEEE– 2002]

(A) e4 (B) e2 (C) e3 (D) e

7.2

xim

p®

l ( )

( )3x22xtan1

xsin12xtan1

-p÷øö

çèæ +

-÷øö

çèæ -

is equal to [AIEEE– 2003]

(A) 161

(B) – 161

(C) 321

(D) – 321

8. If 0xim®

l x

)x3(n)x3(n --+ ll = k , then the value of k is : [AIEEE– 2003]

(A) 0 (B) – 31

(C) 32

(D) – 32


9. Let f(a) = g(a) = k and their nth derivatives fn(a), gn(a) exist and are not equal for some n. Further if

axim®

l )x(f)x(g

)a(g)x(f)a(g)a(f)x(g)a(f-

+-- = 4, then the value of k is equal to [AIEEE– 2003]

(A) 4 (B) 2 (C) 1 (D) 0

10. If ¥®ximl

x2

2xb

xa1 ÷

ø

öçè

æ ++ = e2, then the values of a & b are [AIEEE– 2004]

(A) a Î R, b Î R (B) a = 1, b Î R (C) a Î R, b = 2 (D) a = 1, b = 2

11. Let a and b be the distinct roots of ax2 + bx + c = 0, then 2

2

x )x()cbxaxcos(1im

a-++-

a®l is equal to :

[AIEEE– 2005]

(A) 21

(a – b)2 (B) 2a2

- (a – b)2 (C) 0 (D) 2a2

(a – b)2

12. Let f : R ® R be a positive increasing function with )x(f)x3(flim

x ¥®= 1. Then )x(f

)x2(flimx ¥®

. [AIEEE– 2010]

(A) 32

(B) 23

(C) 3 (D) 1

NCERT BOARD QUESTIONSEvaluate.

1.2

x 3

x 9limx 3®

--

2.2

1x2

4x 1lim2x 1®

--

3.h 0

x h xlimh®

+ -4.

1 13 3

x 0

(x 2) 2limx®

+ -

5.6

2x 0

(1 x) 1lim(1 x) 1®

+ -+ - 6.

5 52 2

x a

(2 x) (a 2)limx a®

+ - +-

7.4

x 1

x xlimx 1®

-

-8.

2

x 2

x 4lim3x 2 x 2®

-

- - +

9.4

2x 2

x 4limx 3 2x 8®

-

+ -10.

7 5

3 2x 1

x 2x 1limx 3x 2®

- +- +

11.3 3

2x 0

1 x 1 xlimx®

+ - - 12.3

5x –3

x 27limx 243®

++


13.2

21x2

8x 3 4x 1lim2x 1 4x 1®

æ ö- +-ç ÷- -è ø 14. Find 'n' , if

n n

x 2

x 2lim 80x 2®

-=

- , n Î N

15.x a

sin3xlimsin7x®

16.2

2x 0

sin 2xlimsin 4x®

17. 2x 0

1 cos2xlimx®

-18. 3x 0

2sin x sin2xlimx®

-

19.x 0

1 cosmxlim1 cosnx®

--

20. x3

1 cos 6xlim

2 x3

p®

-pæ ö-ç ÷

è ø

21. x4

sin x cos xlimx

4p

®

-p

- 22. x6

3 sinx cosxlimx

6p

®

-p

-

23.x 0

sin2x 3xlim2x tan3x®

++

24. x a

sinx sinalimx a®

-

-

25.2

x6

cot x 3limcosec x 2p

®

-- 26. 2x 0

2 1 cosxlimsin x®

- +

27.x 0

sinx 2sin3x sin5xlimx®

- +28. If

4

x 0

x 1limx 1®

--

= 3 3

2 2x k

x klimx k®

--

, then find the value of k.

29. y 0

(x y)sec(x y) xsec xlimy®

+ + -30.

( )x 0

sin( )x sin( )x sin2 xlim .x

cos 2 x cos2 x®

a + b + a - b + a

b - a

31.

3

x4

tan x tanxlimcos x

4p

®

-pæ ö+ç ÷

è ø32. x

x1 sin2lim

x x xcos cos sin2 4 4

®p

-

æ ö-ç ÷è ø

33. Show that x 4

| x 4 |limx 4®

--

does not exists.

34. Let f(x) =

k cosx whenx2x 2

3 x2

pì ¹ïï p -í

pï =ïî

and if x

2

lim f(x)2p

®

pæ ö= ç ÷è ø , find the value of k.

35. Let f(x) = 2

x 2 x 1cx x 1+ £ -ì

í> -î

find 'c' if x 1lim f(x)®-

exists.


EXERCISE # 1PART # I

1. (D) 2. (C) 3. (B) 4. (B) 5. (A) 6. (B) 7. (B)8. (C) 9. (A) 10. (C) 11. (D) 12. (D) 13. (D) 14. (B)15. (B) 16. (D) 17. (C) 18. (A) 19. (C) 20. (C) 21. (A)22. (A) 23. (A) 24. (D) 25. (D) 26. (C) 27. (B) 28. (B)29. (D) 30. (B) 31. (C) 32. (B) 33. (B) 34. (C)

PART # II

1. (i) 2 + sin 2 (ii) tan 3 – 23 (iii)43

cos 43

(iv) 55

2. (i) –18 (ii)

32 (iii)

12 x if x > 0; ¥ if x = 0 (iv) ¥

(v) 2 (vi)32 (vii) 2

3. (i) Limit does not exist (ii) l = –24. (i) No (ii) No (iii) Yes, ¥0 form (iv) No

5. (i) 2516

(ii) 2 (iii) 31

(iv) 2a sina + a2 cos a

6. (i) – 23

(ii) 1912

(iii) 332

7.25

8. a = 2, b = 1, c = –1 and limit = – 31

9. (i) e–1 (ii) 0 (iii) p-

2

e (iv) 1 (v) 1 (vi) 0

10. a + b = 0 and bc = 3

11. (D) 12. (C) 13. (B) 14. (A) 15. (C) 16. (B) 17. (C)

18. (D) 19. (A) 20. (D) 21. (A- S), (B- P) , (C-S) , (D-R)

22. (A- p)(B- p), (C- r) , (D- S) 23. (A- pr), (B-ps), (C-pr), (D- qr)

EXERCISE # 2PART # I

1. (A) 2. (A) 3. (D) 4. (B) 5. (A) 6. (B) 7. (B)

8. (D) 9. (A) 10. (B) 11. (D) 12. (B* 13. (D) 14. (A)

15. (A) 16. (D) 17. (A) 18. (B) 19. (A) 20. (A) 21. (BC)

22. (ABC) 23. (BCD) 24. (A, B, C,D) 25. (A, B) 26. (A, D) 27. (A, B)

28. (A, B, C) 29. (A, B, C, D) 30. (AD)


PART # II

1. (i) 5/4 (ii) 2 (iii) does not exist (iv) does not exist (v) 6

(iv) does not exist (v) does not exist (vi) 8/5 (vii) does not exist

2. (i) 1 (ii) –1/2

3. (2n-1) 4. 9 5. e1

6. (a1 .a2 . a3 ......an)

7. (i) x x

x xx

1 a 1a alim 0 a 1a a 1 0 a 1

-

-®¥

>é- ê= =ê+ ê- < <ë

(ii) x x

x xx

1 a 1a alim 0 a 1a a 1 0 a 1

-

-®-¥

- >é- ê= =ê+ ê < <ë

8. 1/2

9. -cos a 10. 1/2 11. 81

12.12

- 13. 15 14. 1 15. 0

EXERCISE # 3

PART # I

1. (D) 2. (C) 3. (C) 4. (B) 5. (C) 6. (D) 7. (A, C)

8. a = 3, b = 12, c = 9

PART # II

1. (D) 2. (C) 3. (C) 4. (A) 5. (A) 6. (A) 7. (C)

8. (C) 9. (A) 10. (B) 11. (D) 12. (D)

EXERCISE # 4

1. 6 2. 2 3. 2

1x

4.2

31 23

-

5. 3 6. ( )32

5 a 22

+

7. 7 8. 8 9.85 10. 1 11. 0 12.

115

13.72

14. n = 5 15.37

16.14

17. 2 18. 1 19.2

2

mn

20. 3 21. 2 22. 2 23. 1 24. 2 a cosa 25. 4

26.1

4 227. 0 28.

2k3

=

29. sec x (x tan x + 1) 30. 2 2

2aa - b 31. – 4 32.

12

34. k = 6

35. c = 1

limits · a-479 indra vihar, kota rajasthan 324005 ... in the notion of limits note very carefully...

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