$\mathop {\lim }\limits_{x \to {2^ + }} \frac{{1 - \cos \{ {x^2} + 2x\} }}{{\ln {{(x - 1)}^{(x - 2)}}}}$ is equal to (where $\{.\}$ denotes fractional part function).

$\lim _{x \rightarrow \infty} x^3 \left\{\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right\} = $

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{{x^2} + 5x + 3}}{{{x^2} + x + 3}}} \right)^x} = $

The quadratic equation whose roots are $l$ and $m$,where
$\begin{aligned}
& l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right), \\
& m=\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}, \text{ is}
\end{aligned}$

The right hand and left hand limit of the function $f(x)$ are respectively:
$f(x)=\begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text{if } x \neq 0 \\ 0, & \text{if } x=0 \end{cases}$

If $a, b$ and $c$ are three distinct real numbers and $\lim _{x \rightarrow \infty} \frac{(b-c) x^2+(c-a) x+(a-b)}{(a-b) x^2+(b-c) x+(c-a)}=\frac{1}{2}$,then $a+2 c=$