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

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{4}{{x - 1}}} \right)^{3x - 1}} = $

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{{x^2} + 5x + 3}}{{{x^2} + x + 3}}} \right)^x} = $

$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum\limits_{r = 0}^n {{{\tan }^{ - 1}}\left( {1 + r + {r^2}} \right)} }}{n}$ is equal to

Let $\{x\}$ denote the fractional part of $x$ and $f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x \neq 0$. If $L$ and $R$ respectively denote the left-hand limit and the right-hand limit of $f(x)$ at $x=0$,then $\frac{32}{\pi^2}\left(L^2+R^2\right)$ is equal to:

If $\lim _{x \rightarrow \infty}\left(1+\frac{p}{x}\right)^{q x}=e^9$ where $p, q \in \mathbb{N}$,then $p+q=$