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

If $\lim_{x \rightarrow 0} [1 + x \ln(1 + b^2)]^{\frac{1}{x}} = 2b \sin^2 \theta$, where $b > 0$ and $\theta \in (-\pi, \pi]$, then the value of $\theta$ is

Find $\mathop {\lim }\limits_{x \to 1} f(x)$,where $f(x) = \begin{cases} x^{2}-1, & x \leq 1 \\ -x-1, & x > 1 \end{cases}$

$\mathop {\lim }\limits_{m \to \infty } {\left( {\cos \frac{x}{m}} \right)^m} = $

$\mathop {Limit}\limits_{x \to \frac{\pi }{2}} \,\frac{{\sin x}}{{{{\cos }^{ - 1}}\left[ {\frac{1}{4}\,(3\sin x\, - \,\sin 3x)} \right]}}\,$,where $[ \cdot ]$ denotes the greatest integer function,is

Let $[x]$ denote the greatest integer less than or equal to $x$ and $f(x) = 2x - [2x]$. If $\lim_{x \rightarrow 2^{-}} f(x) = l_1$ and $\lim_{x \rightarrow 2^{+}} f(x) = l_2$,then $l_1 + l_2 =$