The value of $\mathop {\lim }\limits_{x \to \infty } \frac{(x + 1)(3x + 4)}{x^2(x - 8)}$ is equal to

$\lim _{x \rightarrow 0} \frac{a^x-1}{\sin (x)} = $

The limit of $\left[\frac{1}{x^{2}}+\frac{(2013)^{x}}{e^{x}-1}-\frac{1}{e^{x}-1}\right]$ as $x \rightarrow 0$ is:

Let $l = \mathop {\lim}\limits_{x \to 0} \frac{[x]^2}{x^2}$ and $m = \mathop {\lim}\limits_{x \to 0} \frac{[x^2]}{x^2}$,where $[ \cdot ]$ denotes the greatest integer function. Then:

If $f: R \rightarrow R$ is defined by $f(x) = [x-3] + |x-4|$ for $x \in R$,then $\lim_{x \rightarrow 3^{-}} f(x)$ is equal to

$\lim _{x \rightarrow \infty} \frac{e^{x^4}-1}{e^{x^4}+1} = $