If $\mathop {\lim }\limits_{x \to 0} \frac{{\log (a + x) - \log a}}{x} + k\mathop {\lim }\limits_{x \to e} \frac{{\log x - 1}}{{x - e}} = 1$,then

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:

If $a > 0$ and $n \in R$,then $\lim_{x \rightarrow a} x^n = \dots$

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 3} \frac{x^{4}-81}{2 x^{2}-5 x-3}$

$\lim _{x \rightarrow \infty} [x - \log (\cosh x)] = $

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} }}{{{x^2}}}$ is equal to