$\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right)}{x^2} = $

Evaluate the limit: $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$

$\lim \limits _{x \to 0} \frac{{{{(\sin x - \tan x)}^2} - {{(1 - \cos 2x)}^4} + {x^5}}}{{7\cdot{{({{\tan }^{ - 1}}x)}^7}\, + {{({{\sin }^{ - 1}}x)}^6}+ 3{{\sin }^5}x}}$ is equal to

If $\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{5 x}=l$ and $\lim _{x \rightarrow 1} \frac{2}{x-1} \log x=m$,then the cubic equation whose roots are $5l, m$,and $1$ is:

$\lim _{x \rightarrow-\infty} \frac{3|x|-x}{|x|-2 x} - \lim _{x \rightarrow 0} \frac{\log (1+x^3)}{\sin ^3 x} =$

The quadratic polynomial $p(x)$ has roots $1$ and $\alpha$, while quadratic polynomial $q(x)$ has roots $1$ and $\beta$. Let $\alpha$ and $\beta$ be the roots of $r(x) = p(x) + q(x)$. Then $\lim_{x \to \infty} [\sqrt{p(x)} - \sqrt{q(x)}] = $