$\mathop {\lim }\limits_{x \to 0} \sin \left( {\frac{1}{x}} \right)$ is

$\lim _{x \rightarrow 0} \frac{(1-e^x) \sin x}{x^2+x^3}$ is equal to

Evaluate the limit: $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$

$\mathop {Lim}\limits_{x \to {0^ - }} \sin^{-1}([\tan x])$ $= l$,then $\{l\}$ is equal to,where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer and fractional part functions,respectively.

$\lim _{x \rightarrow \infty}\left(\frac{x+5}{x+2}\right)^{x+3}$ equals

If $f(x)$ satisfies $97 f(x) + m f\left(\frac{1}{x}\right) = 0$,where $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$ for $x > 0$,then the value of $m$ is: