If $\mathop {\lim }\limits_{x \to \infty } \frac{e^{\mu x} + 5}{e^{100x} + 7}$ exists,then the sum of all possible positive integral values of $\mu$ is:

If $a > 0$,$[\cdot]$ denotes the greatest integer function,$\lim _{x \rightarrow a^{-}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=k$,and $\lim _{x \rightarrow a^{+}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=l$,then:

If $f(x) = \begin{cases} \frac{\sin[x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{cases}$ where $[x]$ denotes the greatest integer less than or equal to $x$,then $\lim_{x \to 0^-} f(x)$ is:

$\mathop {\lim }\limits_{n \to \infty } {({3^n} + {4^n})^{\frac{1}{n}}} = $

$\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2} + 2x - 1}}{{2{x^2} - 3x - 3}} = $

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} - \sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to-