Find the limit: $\mathop {\lim }\limits_{x \to 3} [x(x+1)]$

For each $t \in \mathbb{R}$,let $[t]$ be the greatest integer less than or equal to $t$. Then $\lim_{x \to 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$ is:

$\lim _{x \rightarrow 1} \left( \lim _{y \rightarrow \infty} y \left( (e^x)^{1/y} - 1 \right) \right) = $

$\lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{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:

Find $\mathop {\lim }\limits_{x \to 0} f(x)$ where $f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 0, & x=0 \end{cases}$