\(lim_{x\rightarrow\infty}{sin \ x \over x} \)

(a)

1

(b)

0

(c)

\(\infty\)

(d)

-\(\infty\)

\(lim_{x\rightarrow {\pi/2}}{2x-\pi\over cosx} \)

(a)

2

(b)

1

(c)

-2

(d)

0

\(lim_{x \rightarrow \infty}{\sqrt{x^2-1}\over 2x+1}=\)

(a)

1

(b)

0

(c)

-1

(d)

\(1\over 2\)

\(lim_{x \rightarrow 0}{a^x-b^x\over x}=\)

(a)

log ab

(b)

log\(({a\over b})\)

(c)

log\(({b\over a})\)

(d)

\({a\over b}\)

\(\lim _{ x\rightarrow \infty }{ \left( \frac { 1 }{ x } +2 \right) } \)is equal to

(a)

\(\infty \)

(b)

0

(c)

1

(d)

2

\(\lim _{ x\rightarrow \infty }{ \frac { 1+2+3+....+n }{ { 2n }^{ 2 }+6 } } \)

(a)

2

(b)

6

(c)

\(\frac { 1 }{ 4 } \)

(d)

\(\frac { 1 }{ 2 } \)

\(\lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { \sin { x } }{ x } } =\)

(a)

\(\pi \)

(b)

\(\frac { \pi }{ 2 } \)

(c)

\(\frac { 2 }{ \pi } \)

(d)

1

The point of discontinuity for the function \(\frac { { 2x }^{ 2 }-8 }{ x-2 } \) is

(a)

0

(b)

8

(c)

2

(d)

4

The function y =\(\frac { \left| 3x-4 \right| }{ 3x-4 } \) is discontinuous at x =

(a)

0

(b)

\(\frac { 3 }{ 4 } \)

(c)

\(\frac { 4 }{ 3 } \)

(d)

1

The function \(f\left( x \right) =\tan { x } \) is discontinuous on the set

(a)

\(\left\{ n\pi :\quad n\in z \right\} \)

(b)

\(\left\{ 2n\pi :\quad n\in z \right\} \)

(c)

\(\left\{ (2n+1)\frac { \pi }{ 2 } ,\quad n\in z \right\} \)

(d)

\(\left\{ n\frac { \pi }{ 2 } ,\quad n\in z \right\} \)