$\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\sec \sqrt x } \right)^{\frac{{10}}{x}}}$ is equal to

$\lim _{z \rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1} = $

$\mathop {\lim }\limits_{x \to 0} \cos \frac{1}{x}$

For $\mathop {\text{Lim}}\limits_{x \to 8} \,\,\frac{{\sin \{ x - 10\} }}{{\{ 10 - x\} }}$ (where $\{ \}$ denotes the fractional part function),determine the existence of the limits.

$\lim _{x \rightarrow 0}\left(\frac{\sinh 2 x}{2 x}\right)^{\frac{1}{x^2}} = $

If $\lim _{x \rightarrow 0^{+}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=k$ and $\lim _{x \rightarrow 0^{-}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=l$,then which of the following is true?