Limit ContinuiTY Differential ASSIGNMENT FOR IIT-JEE

L E V E L - 1 (Objective)

1.

cos x − sin x is π   − x  (cos x + sin x ) 4 

lim π x→ 4

(a) 0 2.

(c) e2

(d) none of these

(b) 1

(c) e

(d) does not exist

(b) l n 5

(c) (ln 2) (ln 5)

(d) ln 10

(b) e-2

(c) e-1

(d) 1

10 x − 2 x − 5x + 1 is x tan x

lim x →0

ln 2

 x +1  lim   x →∞ x+2

2 x +1

is

(a) e 6.

(b) e

lim (cos x)cot x is x →0

(a) ln 2 5.

(d) none of these

x →1

(a) 0 4.

(c) -1

lim (log3 3x) logx 3 is

(a) 1 3.

(b) 1

lim

(cos x )1/ 2 − (cos x )1/ 3 is sin 2 x

(a)

1 6

x →0

(b) −

1 12

(c)

2 3

(d)

1 3

1/ x

7.

8.

 a x + bx + cx   lim  x →0 3  

is

(a) abc

(b) abc

π 4

(a) 9.

lim

11 4

(b) x − 2a + x − 2a x 2 − 4a 2

x →2a

(a)

(d) none of these

1 − cot3 x 2 − cot x − cot3 x is

lim x→

(c) (abc)1/3

1 a

(b)

3 4

(c)

1 2

(d) none of these

is 1 2 a

(c)

a 2

(d) none of these

 sin (1 + [ x ]) , [x ] ≠ 0 f ( x ) is 10. If f ( x ) =  [ x ] , then xlim →0 −  0, [x] = 0

(a) -1

(b) 0

11. If f (x ) =

(d) none of these

(c) 0

(d) 1

(c) 1

(d) none of these

sin (e x −2 − 1) , then lim f (x) is x→2 log ( x − 1)

(a) -2

(b) -1

x cos x − log (1 + x) is x2

12. lim x →0 (a)

(c) 1

1 2

(b) 0 πx  =  2

 (1 − x) tan  13. lim x→1

(a) -1

(b) 0

(c)

π 2

(d)

2 π

π − cos−1 x is given by x +1

14. xlim →−1

1 π

(a)

1 2π

(b)

(c) 1

(d) 0

(c) π

(d) none of these

(c) a = 16, b = 8

(d) none of these

(c) 1

(d) none of these

(c) eab

(d) eb/a

(c) − 24

(d)

1 30

(d)

|x +π|

15. xlim → − π sin x is (a) -1 (1+ ax) 16. If lim x→0

(b) 1 b/ x

= e2 , (a, b ∈ N), then

(a) a = 4, b = 2 1 + tan x  17. lim   x →0  1 + sin x 

(a) e

(b) a = 8, b = 4 cos ec x

is (b) e-1

(cosx + a sin bx)1/ x is 18. lim x→0

(a) 1

(a) ab

2 19. If f ( x) = − 25 − x then lim x →1

(a)

1 24

(b)

1 5

 x3  sin x − x + 6  20. Value of lim x →0  x6  

(a) 0

(b)

f ( x) − f (1) is x −1

1 24

   sin x is   

1 12

(c)

1 120

21. Value of lim x →0

1 + sin x − cos x + log(1 − x) is x3

(a) -1/2

(b) 1/2

22. Value of lim x →0

(c) 0

(d) none of these

1 − cos3 x is x sin x cos x

(a)

2 5

(b)

(c)

3 2

(d) none of these

23. If f (x) =

1 18− x

2

, then lim x→3

(b) −

(a) 0

3 5

f (x) − f (3) is x −3

1 9

(c) −

1 3

(d) none of these

1   2  x sin − x  x  is  24. xlim →∞  1− | x |   

(a) 0

(b) 1

(c) -1

(d) none of these

x 2 + 1, x ≠ 0, 2 sin x , x n , n Z ≠ π ∈   x = 0 then lim g(f (x)) is 25. If f ( x ) =  and g(x) =  4, x→0 2 , otherwise   5, x=2 

(a) 5

(b) 4

(c) 2

(d) -5

(b) 0

(c) does not exist

(d) none of these

(b) -1

(c) 0

(d) none of these

(b) 5

(c) 3

(d) none of these

sin [cos x ]

26. lim x →0 1 + [cos x ] is (a) 1 log ( x − a )

27. lim x →a log (e x − e a ) is (a) 1 28. lim x →1

x3 + x 2 − 2 is sin ( x − 1)

(a) 2

1 − cos λx , x≠0  x 29. If f ( x ) =  x sin is continuous at x = 0, then λ is 1  , x =0 2 

(b) ±1

(a) 0

(c) 1

(d) none of these

[ x ] + [− x ], x ≠ 2 , then 'f' is continuous at x = 2 provided λ is x=2 λ, 

30. If f ( x ) =  (a) -1

(b) 0

(c) 1

(d) 2

31. If f (x) =

2 − (256− 7x)1/ 8 , (x ≠ 0) then for 'f' to be continuous everywhere f(0) is equal to (5x + 32)1/ 5 − 2

(a) -1

(b) 1

(c) 16

 x2  a ,  32. The function f ( x ) =  a ,  2b 2 − 4b ,  2  x

(d) none of these

0 ≤ x