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

(a)

1

(b)

0

(c)

\(\infty\)

(d)

-\(\infty\)

\(lim_{x\rightarrow0}{\sqrt{1-cos 2x}\over x} \)

(a)

0

(b)

1

(c)

\(\sqrt{2}\)

(d)

does not exist

If f(x) = x(-1)^{\(\left\lfloor 1\over x \right\rfloor \)}, \(x\le0\), then the value of \(lim_{x\rightarrow 0}f(x)\) is equal to

(a)

-1

(b)

0

(c)

2

(d)

4

Let the function f be defined by \(f(x)=\left\{\begin{array}{ll} 3 x & 0 \leq x \leq 1 \\ -3 x+5 & 1<x \leq 2 \end{array},\right. \text { then }\)

(a)

\(lim_{x \rightarrow1}f(x)=1\)

(b)

\(lim_{x \rightarrow1}f(x)=3\)

(c)

\(lim_{x \rightarrow1}f(x)=2\)

(d)

\(lim_{x \rightarrow1}f(x)\) does not exist

The value of \(lim_{x \rightarrow 0}{sin x\over \sqrt{x^2}}\) is

(a)

1

(b)

-1

(c)

0

(d)

limit does not exist

The function \(f(x)= \begin{cases}\frac{x^{2}-1}{x^{3}+1} & x \neq-1 \\ P & x=-1\end{cases}\)is not defined for x = −1. The value of f(−1) so that the function extended by this value is continuous is

(a)

\({2\over3}\)

(b)

-\({2\over3}\)

(c)

1

(d)

0

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

(a)

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

(b)

2

(c)

\(\frac { 11 }{ 4 } \)

(d)

0

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

(a)

\(\infty \)

(b)

0

(c)

1

(d)

2

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

(a)

\(\pi \)

(b)

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

(c)

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

(d)

1

The slop of the curve \(y={ x }^{ 3 }-7{ x }^{ 2 }+5x+10\) at x=1 is

(a)

6

(b)

-6

(c)

5

(d)

15