Let $[P]$ denote the greatest integer $\leq P$. If $0 \leq a \leq 2$,then the number of integral values of $a$ such that $\lim _{x \rightarrow a}([x^2]-[x]^2)$ does not exist is:

Let $a_{1}, a_{2}, \dots, a_{n}$ be fixed real numbers and define a function $f(x) = (x - a_{1})(x - a_{2}) \dots (x - a_{n})$. What is $\lim_{x \to a_{1}} f(x)$? For some $a \neq a_{1}, a_{2}, \dots, a_{n}$,compute $\lim_{x \to a} f(x)$.

Let $[x]$ denote the greatest integer less than or equal to $x$. Then,evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \,\frac{{\tan \,(\pi \,{{\sin }^2}\,x) + \,{{(\left| x \right|\, - \,\sin \,(x\,[x]))}^2}}}{{{x^2}}}$

Find the limit: $\mathop {\lim }\limits_{x \to -1} (1+x+x^{2}+ \dots +x^{10})$

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:

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $