The value of $\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\sin \frac{1}{x} - x}}{{1 - |x|}}$ is

Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right-hand limit $\lim_{x \rightarrow 0^{+}} \frac{(1-x)^{\frac{1}{x}}-e^{-1}}{x^a}$ is equal to a nonzero real number is:

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x = $

The value of $\lim _{x \rightarrow 0} \left( \frac{1}{x} \ln \sqrt{\frac{1+x}{1-x}} \right)$ is

$[x]$ denotes the greatest integer less than or equal to $x$. If $\{x\}=x-[x]$ and $\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x+[x])}{2-\{x\}}=\theta$,then $\sin \theta+\cos \theta=$

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: