Limits and Cont...

Let $$f\left ( x \right )=\begin{cases}\sin x, x\neq n\pi 
                   \\ 2,  x=n\pi \end{cases}$$, where $$n\epsilon \mathbb{Z}$$ and
$$g\left ( x \right )=\begin{cases}x^{2}+1, x\neq 2 \\
              3, x=2 \end{cases}$$.
Then $$\displaystyle \lim_{x\to 0}g\left ( f\left ( x \right ) \right )$$ is

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\left ( 1-\tan \dfrac{x}{2} \right )\left ( 1-\sin x \right )}{\left ( 1+\tan \dfrac{x}{2} \right )\left ( \pi -2x \right )^{3}}$$ is

Evaluate $$\displaystyle \lim_{n\rightarrow \infty }\left [ \frac{n!}{n^{n}} \right ]^{1/n}$$.

$$\displaystyle \lim_{x\rightarrow\infty}\left(\frac{\sqrt{(1 - \cos x)+ \sqrt{(1 - \cos x)+ \sqrt(1 - \cos x)+...\infty) - 1}}}{x^2}\right)$$ equals to

Evaluate $$\displaystyle \lim_{n\rightarrow \infty }\left [ \left ( 1+\frac{1}{n^{2}} \right )\left ( 1+\frac{2^{2}}{n^{2}} \right )\left ( 1+\frac{3^{2}}{n^{2}} \right )......\left ( 1+\frac{n^{2}}{n^{2}} \right ) \right ]^{1/n}$$

$$\underset{x\rightarrow0}{lim}\displaystyle\frac{1-cos^{3}x+sin^{3}x+\ell n(1+x^{3})+\ell n(1+cos\,\,x)}{x^{2}-1+2\,cos^{2}x+tan^{4}x+sin^{3}x}$$ is equal to -

let a, b, c are non zero constant number then $$\lim_{r\rightarrow\infty}\displaystyle\frac{cos\displaystyle\frac{a}{r}-cos\displaystyle\frac{b}{r}cos\displaystyle\frac{c}{r}}{sin\displaystyle\frac{b}{r}sin\displaystyle\frac{c}{r}}$$ equals to

The limit of $$x\sin { \left( { e }^{ \frac { 1 }{ x }  } \right)  } $$ as $$x\rightarrow 0$$

If $$\displaystyle \lim_{x\rightarrow \infty}\dfrac{x^3+1}{x^2+1}-(ax+b)=2$$, then

$$\displaystyle \lim_{x \rightarrow \infty} (\sqrt{x^2 + 8x + 3} - \sqrt{x^2 + 4x + 3}) =$$