$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}} = $

$\lim _{x \rightarrow 0} \frac{x^2 \log (\cos x)}{\log (1+x^2)} = $

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} \frac{1}{n^3} \sum_{k=1}^n k^2 x = $

If $x$ is a real number in $[0, 1]$,then the value of $\lim_{m \to \infty} \lim_{n \to \infty} [1 + \cos^{2m}(n! \pi x)]$ is given by