Evaluate the given limit: $\mathop {\lim }\limits_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sum_{k=1}^{n} {k^2}}}{{{n^3}}}} \right] = $

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }}$ is equal to

Let $f(x)$ be a differentiable function such that $f(0)=0$ and $f^{\prime}(0)=20$. For $x \in \left(0, \frac{\pi}{2}\right]$,if $A(x)=2 f(x) \operatorname{cosec} 4 x+4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x$,then $\lim _{x \rightarrow 0} A(x)=$

If $f(x) = \begin{cases} x & \text{if } x < 0 \\ 1 & \text{if } x = 0 \\ x^2 & \text{if } x > 0 \end{cases}$,then $\mathop {\lim }\limits_{x \to 0} f(x) = $

$\lim _{n \rightarrow \infty} n\left(\sqrt{n^2+9}-n\right)=$