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

$\lim _{x}$ ${\rightarrow 0^{+}} \frac{\tan \left(5(x)^{\frac{1}{3}}\right) \log _e\left(1+3 x^2\right)}{\left(\tan ^{-1} 3 \sqrt{x}\right)^2\left(e^{5(x)^{\frac{4}{3}}}-1\right)}$ is equal to

If $\mathop {\lim }\limits_{x \to 2} \frac{{{x^n} - {2^n}}}{{x - 2}} = 80$,where $n$ is a positive integer,then $n = $

If $f(x) = \cos x$ when $x = n\pi$ $(n = 0, 1, 2, 3, \dots)$ and $f(x) = 3$ otherwise,and $\phi(x) = \begin{cases} x^2 + 1 & \text{when } x \neq 3, x \neq 0 \\ 3 & \text{when } x = 0 \\ 5 & \text{when } x = 3 \end{cases}$,then find $\lim_{x \to 0} f(\phi(x))$.

Consider the following statements:
$I$. $\lim _{n \rightarrow \infty} \frac{2^n+(-2)^n}{2^n}$ does not exist.
$II$. $\lim _{n \rightarrow \infty} \frac{3^n+(-3)^n}{4^n}$ does not exist.
Then,

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{\cos x}{\pi - x}$