$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{{\sin }^2}x}}$ is equal to

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:

$\lim _{n \rightarrow \infty}\left\{n-\sqrt{n^2-4 n}\right\}=$

Let for all $x > 0$, $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$, then

$\mathop {\lim }\limits_{x \to 0} {(1 - ax)^{\frac{1}{x}}} = $

$\lim _{x \rightarrow \infty}\left(\frac{x+5}{x+2}\right)^{x+3}$ equals