Limits and Cont...

$$\displaystyle \lim_{n\rightarrow \infty }\left(\displaystyle \frac{e^{n}}{\pi}\right)^{1/n}=$$

$$\displaystyle \lim_{x\rightarrow 1}(2-x)^{\displaystyle \tan( \frac{\pi x}{2})}=$$

lf the function defined by $$\mathrm{f}({x})=\displaystyle \frac{\sin 3(x-p)}{\sin 2(x-p)}$$ for $${x}\neq {p}$$ is continuous at $${x}={p}$$ then $$\mathrm{f}({p})=$$

If the function $$\displaystyle \mathrm{f}({x})=\begin{cases}\dfrac{2^{x+2}-16}{4^{x}-16}&&  for {x}\neq 2\\ \mathrm{A} && x =2\end{cases}$$ is continuous at $$x =2$$, then $$\mathrm{A}=$$

 $$f(x)=x\left[3-\displaystyle \log\left(\frac{\sin {x}}{x}\right)\right]-2$$ to be continuous at $${x}=0$$, then $$\mathrm{f}({0})=$$

 Let $$\displaystyle \mathrm{f}({x})=\begin{cases} \dfrac{(e^{kx}-1).\sin kx}{x^{2}} & for \ {x}\neq 0   \\ 4 & for \ {x} =0\end{cases}$$ is continuous at $${x}=0$$ then $${k}=$$

 lf $$f(x)=
\left\{\begin{matrix} (1+|\sin x|)^{\displaystyle \frac{a}{|\sin x|}}&-\displaystyle \frac{\pi}{6}<x<0\\  b&x=0 \\ e^{\displaystyle \frac{\tan 2x}{\tan 3x}} &0<x<\displaystyle \frac{\pi}{6}\end{matrix}\right.$$ is

continuous at $$\mathrm{x}=0$$ then

lf the function $$f(x)=\begin{cases}\dfrac{k\cos x}{\pi-2x}, & x\neq\dfrac{\pi}{2}\\ 3 & at x=\dfrac{\pi}{2}\end{cases}$$is continuous at $$\displaystyle {x}=\dfrac{\pi}{2}$$ then $${k}=$$

The function $$f(x)=\begin{cases} 0,&  \text{x  is irrational }\\  1,& \text{x is rational }\end{cases}$$ is

The function $$\displaystyle \mathrm{f}(\mathrm{x})=\frac{\cos x-\sin x}{\cos 2x}$$ is not defined at $$x=\displaystyle \frac{\pi}{4}$$ The value of $$f\left(\displaystyle \frac{\pi}{4}\right)$$ so that $$\mathrm{f}(\mathrm{x})$$ is continuous at $$x=\displaystyle \frac{\pi}{4}$$ is