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

If $f(x) = \begin{cases} x^2 - 3, & 2 < x < 3 \\ 2x + 5, & 3 < x < 4 \end{cases}$,the equation whose roots are $\lim_{x \to 3^-} f(x)$ and $\lim_{x \to 3^+} f(x)$ is

$\mathop {Lim}\limits_{x \to c} f(x)$ does not exist when: (where $[x]$ denotes the greatest integer function and $\{x\}$ denotes the fractional part function.)

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} ) = $

The limit $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$

$\mathop {\lim }\limits_{x \to \pi /2} \frac{\tan 3x}{x} = $