Limits and Cont...

$$f(x)=\displaystyle \frac{e^{1/x^{2}}}{e^{1/x^{2}}-1}$$ , $$x\neq 0$$, $$\mathrm{f}({0})=1$$, then $$\mathrm{f}$$ at $${x}=0$$ is:

 $$\mathrm{A}$$ function $$\mathrm{f}(\mathrm{x})$$ is defined as
$$f(x)=\left\{ \begin{matrix} ax-b & x\leq 1 \\ 3x, & 1<x<2 \\ bx^{ 2 }-a & x\geq 2 \end{matrix} \right. $$ is continuous at
$$x=1, 2$$ then:

$$Lt_{x \rightarrow 0}\dfrac{\sin 2x+a\sin x}{x^{3}}$$ exists and finite then $$\mathrm{a}=$$

The integer $$n$$ for which $$\displaystyle \lim_{x\rightarrow 0}\frac{(\cos x-1)(\cos x-e^{x})}{x^{n}}$$ is finite non zero number is

$$\displaystyle \lim_{x\rightarrow 1}\{1-x+[x+1]+[1-x]\}$$ , where $$[x]$$ denotes greatest integer function, is

Given that the function $$\mathrm{f}$$ is defined by $$f(x)=\left\{\begin{array}{l}2x-1,x>2\\k, x=2\\x^{2}-1,x<2\end{array}\right.$$is continuous at x = 2. Then $${k}$$ is:

lf the function $$\mathrm{f}({x})=\begin{cases}\dfrac{\sin 3x}{x} &(x\neq 0) \\ \dfrac{k}{2}&(x=0) \end{cases}$$ is continuous at $${x}=0$$, then $${k}$$ is:

$$\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}\cos ^{ -1 }{ \left( \frac { 1-x^{ 2 } }{ 1+x^{ 2 } }  \right)  } =$$

The right-hand limit of the function $$\sec{x}$$ at $$\displaystyle x=-\frac { \pi  }{ 2 } $$ is

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1+a^{3})+8e^{1/x}}{1+(1-b^{3})e^{1/x}}=2$$ then