limit contion differ

8/14/2019 Limit Contion Differ

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CONTENTS

* Synopsis

Questions

* Level - 1

* Level - 2

* Level - 3

Answers

* Level - 1

* Level - 2

* Level - 3

e-Learning Resources

w w w . m a t h i i t . i n

LIMITSUNIT - 2

CONTINUITY

DIFFERENTIABILITY


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Mathiit e- Le arn in g Res our ce [email protected] www.mathiit.in

(f) Limitx a

x a

x an a

n nn

= 1 .

General Note :

All the above standard limits can be generalized if'x

' is replaced by f (x) .

e.g. Limitx asin ( )

( )

f x

f x = 1 provided Limitx a f(x) = 0 & so on .

4. Indeterminant Forms :

1and,,0,0,,0

0 00 .

Note:

(i) Note here that '0

' doesn't means exact zero but represent a value approaching towards zero

similarly for '1' and infinity .(ii) + =

(iii) = (iv) (a/ ) = 0 if a is finite

(v)a

0is not defined for any a R.

(vi) ab = 0 , if and only if a = 0 or b = 0 and a & b are finite .5. To evaluate a limit, we must always put the value where 'x' is approaching to in the function. If we

get a determinate form, than that value becomes the limit otherwise if an indeterminant form comes,then apply one of the following methods :(a)Factorisation

(b)Rationalisation or double rationalisation(c)Substitution(d)Using standard limits(e)Expansion of functions.The following expansions must be remembered :

(i) 0a............!3

anx

!2

anx

!1

anx1a

3322x >++++=

lll

(ii) ex x xx = + + + +11 2! 3

2 3

! !

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

(iii) ln (1+x) = xx x x

for x + + < 2 3 4

2 3 41 1.........

(iv) sin!

.......x xx x x

= + +3 5 7

3 5! 7!

(v) cos! !

......xx x x

= + +12! 4 6

2 4 6

(vi) tan x = xx x

+ + +

3 5

3

2

15 ........


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(vii) tan-1x = xx x x

+ +3 5 7

3 5 7.......

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

3

1 3

5!

1 3 5

7!

23

2 25

2 2 27

!

. . ........

(ix) sec-1x = 12!

5

4

61

6

2 4 6

+ + + +x x x

! !......

1. A function f(x) is said to be continuous at x = c, if Limitx c

f(x) = f(c) . Symbolically 'f' is

continuous at x = c if Limith 0

f(c - h) = Limith 0

f(c + h) = f(c).

i.e. LHL at x = c = RHL at x = c equals value of f at x = c.2. A function 'f' can be discontinuous due to any of the following three reasons :

(i) Limitx c

f(x) does not exist i.e. Limitx c

f(x) Limitx c +

f (x)

(ii) f(x) is not defined at x = c

(iii)Limitx c

f(x) f (c)

Geometrically, the graph of the function will exhibit a break at x = c .

3. (a) In case Limitx c f(x) exists but is not equal to f(c) then the function is said to have a removable

discontinuity or discontinuity of the first kind . In this case we can redefine the function such that

Limitx c

f(x) = f(c) & make it continuous at x = c .

Removable type of Discontinuity can be further classified as :

(i)Missing Point Discontinuity : Where Limitx a

f(x) exists finitely but f(a) is not defined .

e.g. f(x) =( ) ( )

( )

1 9

1

2

x x

xhas a missing point discontinuity at x = 1 .

(ii)Isolated Point Discontinuity : Where Limitx a

f(x) exists & f(a) also exists but,

Limitx a

f(x) f(a) . e.g. f(x) = xx

2 16

4

, x 4 & f (4) = 9 has a break at x = 4 .

(b) In case Limitx c

f(x) does not exist then it is not possible to make the function continuous by

redefining it. However if both the limits (i.e.L.H. L. & R.H.L.) are finite, then discontinuity is saidto be of first kind otherwise it is non-removable discontinuity of second kind .

CONTINUITY


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Irremovable type of Discontinuity can be further classified as :

(i)Finite discontinuity: e.g. f(x) = x - [x] at all integral x.

(ii)Infinite discontinuity: e.g. f(x) =1

4x or g(x) =

1

4 2( )x at x = 4.

(iii)Oscillatory discontinuity: e.g. f(x) = sin 1

x

at x = 0.

In all these cases the value of f(a) of the function at x = a (point of discontinuity) may or may not

exist but Limitx a

does not exist .

4. In case of non-removable discontinuity of the first kind the non-negative difference between thevalue of the RHL at x = c & LHL at x = c is called "The jump of Discontinuity". A functionhaving a finite number of jumps in a given interval I is called a "Piece wise Continuous" or"Sectionally Continuous" function in this interval .

5. All polynomials, trigonometrical functions, exponential & logarithmic functions are continuous in

their domains .6. If f & g are two functions that are continuous at

x = c then the functions defined by :F

1(x) = f(x) g(x) ; F

2(x) = Kf(x) , K any real number ; F

3(x) = f(x).g(x) are also continuous

at x = c. Further, if g (c) is not zero, then F4(x) =

f x

g x

( )

( )is also continuous at x = c .

Note Carefully:

(a) If f(x) is continuous & g(x) is discontinuous at

x = a then the product function (x) = f(x).g(x)

may be continuous but sum or difference function (x) = f(x)+ g(x) will necessarily be discontinuous

at x = a . e.g.

f (x) = x & g(x) =sin x x

x

=

0

0 0

(b) If f (x) and g(x) both are discontinuous at x = a then the product function (x) = f(x).

g(x) is not necessarily be discontinuous at x = a .

e.g. f(x) = -g(x) =1 0

1 0

x

x

>

xO

y

a

Figure 1

Figure 2

>

>

xO

y

a

>

>

xO

y

a

Figure 3

>

>

xO

y

a

Figure 4



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1.)xsinx(cosx

4

xsinxcoslim

4x +

is

(a) 0 (b) 1 (c) -1 (d) none of these

2.3log

31x

x)x3(loglim

is

(a) 1 (b) e (c) e2 (d) none of these

3.

xcot

0x )x(coslim is(a) 0 (b) 1 (c) e (d) does not exist

4.xtanx

15210lim

xxx

0x

+ is

(a) ln 2 (b)5nl

2nl(c) (ln 2) (ln 5) (d) ln 10

5.1x2

x 2x

1xlim

+

++ is

(a) e (b) e-2 (c) e-1 (d) 1

6.xsin

)x(cos)x(coslim

2

3/12/1

0x

is

(a)6

1(b)

12

1 (c)

3

2(d)

3

1

7.x/1

xxx

0x 3

cbalim

++

is

(a) abc (b) abc (c) (abc)1/3 (d) none of these

8. xcotxcot2xcot1

lim3

3

4x

is

(a)4

11(b)

4

3(c)

2

1(d) none of these

9.22a2x a4x

a2xa2xlim

+

is

(a)a

1 (b)a2

1 (c)2

a(d) none of these

w w w . m a t h i i t . i nLEVEL - 1 (Objective)



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10. If

=

+

=0]x[,0

0]x[,]x[

])x[1(sin

)x(f , then )x(flim0x

is

(a) -1 (b) 0 (c) 1 (d) none of these

11. If )1x(log

)1e(sin)x(f

2x

=

, then )x(flim2x

is

(a) -2 (b) -1 (c) 0 (d) 1

12. 20x x)x1(logxcosx

lim+

is

(a)2

1(b) 0 (c) 1 (d) none of these

13. =

2

xtan)x1(lim

1x

(a) -1 (b) 0 (c) 2

(d)

2

14.1x

xcoslim

1

1x +

is given by

(a)

1(b)

21

(c) 1 (d) 0

15.xsin

|x|lim

x

+ is

(a) -1 (b) 1 (c) (d) none of these

16. If ),Nb,a(,e)ax1(lim2x/b

0x=+

then

(a) a = 4, b = 2 (b) a = 8, b = 4 (c) a = 16, b = 8 (d) none of these

17.xeccos

0x xsin1

xtan1lim

++

is

(a) e (b) e-1 (c) 1 (d) none of these

18.x/1

0x)bxsinax(coslim +

is

(a) 1 (a) ab (c) eab (d) eb/a

19. If 1x

)1(f)x(flimthenx25)x(f

1x

2

=

is

(a)24

1(b)

5

1(c) 24 (d) 24

1

20. Value of xsinx

6

xxxsin

lim6

3

0x

+

is

(a) 0 (b)12

1(c)

30

1(d)

120

1



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21. Value of 30x x)x1log(xcosxsin1

lim++

is

(a) -1/2 (b) 1/2 (c) 0 (d) none of these

22. Value ofxcosxsinx

xcos1lim

3

0x

is

(a)52 (b)

53

(c)2

3(d) none of these

23. If 3x

)3(f)x(flimthen,

x18

1)x(f

3x2

=

is

(a) 0 (b)9

1 (c)

3

1 (d) none of these

24.

|x|1

xx

1sinx

lim

2

xis


25. If

=otherwise,2

Zn,nx,xsin)x(f and

==

+=

2x,5

0x,4

2,0x,1x

)x(g

2

then ))x(f(glim0x

is

(a) 5 (b) 4 (c) 2 (d) -5

26.]x[cos1

]x[cossinlim

0x + is

(a) 1 (b) 0 (c) does not exist (d) none of these

27.)ee(log

)ax(loglim

axax

is

(a) 1 (b) -1 (c) 0 (d) none of these

28. )1x(sin2xx

lim

23

1x

+ is

(a) 2 (b) 5 (c) 3 (d) none of these

29. If

=

=0x,

2

1

0x,xsinx

xcos1

)x(f is continuous at x = 0, then is

(a) 0 (b) 1 (c) 1 (d) none of these

30. If

=

+=

2x,

2x],x[]x[)x(f , then 'f' is continuous at x = 2 provided is

(a) -1 (b) 0 (c) 1 (d) 2



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31. If )0x(,2)32x5(

)x7256(2)x(f

5/1

8/1

+

= then for 'f' to be continuous everywhere f(0) is equal to

(a) -1 (b) 1 (c) 16 (d) none of these

32. The function


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39. Let

+

>+= 4x,8x2

4x,dy|)2y|3()x(f

x

0then

(a) f(x) is continuous as well as differentiable everywhere.(b) f(x) is continuous everywhere but not differentiable at x = 4

(c) f(x) is neither continuous nor differentiable at x = 4. (d) 2)4(fL =

40. If f(x) = cos(x2 - 2[x]) for 0 < x < 1, where [x] denotes the greatest integer x , then

2

f is equal to

(a) (b) (c)2

(d) none of these

41. The following functions are continuous on ),0(

(a) tan x (b) x

0

dtt

1sint (c)


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1. Let2

)y(f)x(f

2

yxf

+=

+ for all real x and y. If )0(f exists and equals to -1 and f(0) = 1, then f(2) is equal

to...

2. If 0)0(fandn

1nflim)0(fandR]1,1[:f

n=

=

. Then the value of n

n

1cos)1n(

2lim

1

n

+

is ........

Given that 2n1coslim0 1

n +

=

==

++ =


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9. Let

==

+

0x,0

0x,xe)x(fx

1

|x|

1

Test whether(a) f(x) is continuous at x = 0 (b) f(x) is differentiable at x = 0.

10. If

+=+ xy1

yxf)y(f)x(f for all x, )1xy(,Ry and 2x

)x(flim0x = . Find

31f and )1(f .

11. (a) If f(2) = 4 and f(2) = 1, then find2x

)x(f2)2(xflim2x

(b) f(x) is differentiable function given

f(1) = 4, f(2) = 6, where f(c) means the derivative of function at x = c then)1(f)hh1(f

)2(f)hh22(f2

2lim

0h +++

.

12.


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20 A function 'f' is defined as f(x) =

1 12

12

2

| || |

| |

xif x

a x b x c if x

+ + 0)

25 If f (x) =4x

5xsinxBxcosA +(x 0) is continuous at x = 0, then find the value of A and B . Also find f(0)

. Use of series expansion or L Hospital's rule is prohibited.



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1. Let and be the distinct roots of ax2 + bx + c = 0, then 22

x )x(

)cbxax(cos1lim

++

is equal to

(a) 22

)(2

a

(b) 2)(

2

1

(c) 22

)(2

a (d) 0

2. If ,0x,x

1sinx)x(f

= then =

)x(flim

0x

(a) 1 (b) 0 (c) -1 (d) does not exist

3. =+

20x x

)x1log(xcosxlim

(a)2

1(b) 0 (c) 1 (d) none of these

4. =

1xcot

1xcos2

lim4

x

(a)2

1(b)

2

1(c)

22

1(d) 1

5. =+

1x1

1alim

x

0x

(a) 2 logea (b)

2

1log

ea (c) a log

e2 (d) none of these

6. =

2

x

0x x

xcoselim

2

(a)2

3(b)

2

1(c)

3

2(d) none of these

7.

+


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8. =

x3sinx

x5sin)x2cos1(lim

20x

(a)3

10(b)

10

3(c)

5

6(d)

6

5

9.

=

=2x,k2x,2x

32x

)x(f

5

is continuous at x = 2, then k =

(a) 16 (b) 80 (c) 32 (d) 8

10. If a, b, c, d are positive, then =

++

+

dxc

x bxa

11lim

(a) ed/b (b) ec/a (c) e(c+d)/(a+b) (d) e

11. =

xxtan

eelim

xxtan

0x

(a) 1 (b) e (c) e - 1 (d) 0

12. The value of k which makes

=

=0x,k

0x,x

1sin

)x(f continuous at x = 0 is


13. The value of f(0) so that the functionxtanx2

xsinx2)x(f

1

1

+

= is continuous at each point on its domain is

(a) 2 (b)31 (c)

32 (d)

31

14. If the function

=

++=

2xforx

2xfor2x

Ax)2A(x)x(f

2

is continuous at x = 2, then

(a) A = 0 (b) A = 1 (c) A = -1 (d) none of these

15. Letx

xsin1xsin1)x(f

+= . The value which should be assigned to 'f' at x = 0 so that it is continuous

everywhere is

(a)2

1(b) -2 (c) 2 (d) 1

16. The number of points at which the function|x|log

1)x(f = is discontinuous is

(a) 1 (b) 2 (c) 3 (d) 4

17. The value of 'b' for which the function


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18. The value of f(0) so that

+

=

3

x1log

4

xsin

)14()x(f

2

3x

is continuous everywhere is

(a) 3(ln 4)3 (b) 4 (ln 4)3 (c) 12 (ln 4)3 (d) 15 (ln 4)3

19. Let


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28. If ,kx

)x3(log)x3(loglim

0x=

+

then k is

(a)3

1 (b)

3

2(c)

3

2 (d) 0

29. If ,ex

b

x

a1lim 2

x2

2

x

=

++

then the values of 'a' and 'b' are

(a) Rb,Ra (b) Rb,1a =

(c) 2b,Ra = (d) a = 1 and b = 2

30. If xcosx

xsinx)x(f

2+

= , then )x(flimx

is

(a) 0 (b) (c) 1 (d) none of these

31. =

+

1)x1(

12lim

2/1

x

0x

(a) log 2 (b) 2 log 2

(c) 2log2

1(d) 0

32. =

30x x

xsinxtanlim

(a) 1 (b) 2 (c)2

1(d) -1

33. The function

limit contion differ

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