$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to

$\lim _{n \rightarrow \infty}\left[\frac{1^3}{1-n^4}+\frac{2^3}{1-n^4}+\ldots +\frac{n^3}{1-n^4}\right]=$

Let $[x]$ denote the greatest integer not exceeding $x$. If $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$,$l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$,then:

If $l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^3 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \left( \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} \right)$,find the quadratic equation whose roots are $l$ and $m$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos (1 - \cos x)}}{{x\tan x - {x^2}}}$ is:

$\lim _{x \rightarrow 0}\left(\frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}\right)^{\frac{100}{x}}$ is equal to $.....$