$\mathop {\lim }\limits_{x \to 0} \cos \frac{1}{x}$

For the function $f(x) = \lim_{n \to \infty} \frac{1}{1 + n \sin^2(\pi x)}$,which of the following holds?

$\mathop {\lim }\limits_{x \to 0} \left( \frac{x}{\tan^{-1} 2x} \right) = $

Given below are two statements:
Statement $I$: $\lim _{x \rightarrow 0} \left( \frac{\tan ^{-1} x + \log _e \sqrt{\frac{1+x}{1-x}} - 2x}{x^5} \right) = \frac{2}{5}$
Statement $II$: $\lim _{x \rightarrow 1} \left( x^{\frac{2}{1-x}} \right) = \frac{1}{e^2}$
In the light of the above statements,choose the correct answer from the options given below:

$\lim _{x \rightarrow 0}\left[\tan \left(x+\frac{\pi}{4}\right)\right]^{1 / x}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{x\sqrt {{y^2} - {{(y - x)}^2}} }}{{{{(\sqrt {8xy - 4{x^2}} + \sqrt {8xy} )}^3}}}$ is equal to