The value of $\mathop {\text{Limit}}\limits_{x \to 0} \frac{\tan(\{x\} - 1) \sin\{x\}}{\{x\}(\{x\} - 1)}$,where $\{x\}$ denotes the fractional part function,is:

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos (1 - \cos x)}}{{x\tan x - {x^2}}}$ is:

The value of $\lim _{n}$ ${\rightarrow \infty}\left[\left(\frac{1}{2 \cdot 3}+\frac{1}{2^2 \cdot 3}\right)+\left(\frac{1}{2^2 \cdot 3^2}+\frac{1}{2^3 \cdot 3^2}\right)+\ldots+\left(\frac{1}{2^n \cdot 3^n}+\frac{1}{2^{n+1} \cdot 3^n}\right)\right]$ is

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 _{x \rightarrow-\infty} \frac{3|x|^3-x^2+2|x|-5}{-5|x|^3+3 x^2-2|x|+7} = $

If $g(x) = \frac{x}{[x]}$ for $x > 2$,then $\lim_{x \rightarrow 2^+} \frac{g(x) - g(2)}{x - 2}$ is equal to