For $x \in R$,$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x - 3}}{{x + 2}}} \right)^x}$ is equal to

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:

Let $f(x) = \frac{\sqrt{x+3}}{x+1}$. Then the value of $\lim_{x \rightarrow -3^{-}} f(x)$ is

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \cdot \frac{1-\sin x}{(\pi-2 x)^3} = $

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

Find the limit: $\mathop {\lim }\limits_{x \to -1} (1+x+x^{2}+ \dots +x^{10})$