$\underset{x\to \frac{\pi }{2}}{lim}\frac{\cot x-\cos x}{(\pi -2x{)}^3}$
Let $x=\frac{\pi }{2}+t$
If $x\to \frac{\pi }{2},t\to 0$
$\underset{t\to 0}{lim}\frac{\sin t-\tan t}{-8t^3}$
$=\underset{t\to 0}{lim}\frac{(t-\frac{t^3}{3!}+\frac{t^5}{5!}+\dots )-(t+\frac{t^3}{3}+\frac{2t^5}{15}+\dots )}{-8t^3}$
$=\frac{1}{16}$
We can put  $x=\frac{\pi }{2}-t$ and we'll get L.H.L also same.
Since L.H.L = R.H.L the limit exists and is $=\frac{1}{16}$

$\underset{x\to -\infty }{lim}\left[\frac{x^4\sin \left(\frac{1}{x}\right)+x^2}{1+|x{|}^3}\right]$
$\underset{x\to -\infty }{lim}\frac{x^3}{1+|x{|}^3}[x\sin \left(\frac{1}{x}\right)+\frac{1}{x}]$
As $x\to -\infty$
$\frac{x^3}{1+|x{|}^3}=-1$
$xsin\left(\frac{1}{x}\right)+\frac{1}{x}=1$
$\Rightarrow \underset{x\to -\infty }{lim}\frac{x^3}{1+|x{|}^3}(xsin\frac{1}{x}+\frac{1}{x})=-1$

Given that f(9) = 9, f’(9) = 4
$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}$
On rationalisation we get - 
$=\underset{x\to 9}{lim}\left(\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}\right)\left(\frac{\sqrt{x}+3}{\sqrt{x}+3}\right)\left(\frac{\sqrt{f\left(x\right)}+3}{\sqrt{f\left(x\right)}+3}\right)$
$=\underset{x\to 9}{lim}\frac{f\left(x\right)-f\left(9\right)}{x-9}$ . $\frac{6}{6}$
$=\underset{x\to 9}{lim}\frac{f\left(x\right)-f\left(9\right)}{x-9}$
$=f^{{}^{\prime }}\left(9\right)$
$=4$

$\because \underset{x\to a}{lim}\frac{x^3-a^3}{x-a}=(\frac{0}{0}\text{inderminant form)}$

$\underset{x\to a}{lim}sinxcosecx=(0\times \infty \text{inderminant form)}$

$\underset{x\to a}{lim}{x}^x=(0^0\text{inderminant form)}$

$\underset{x\to a}{lim}\frac{0}{x}=0(\text{not inderminant form)}$