If $a = \lim_{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ and $b = \lim_{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$,then

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} - \sqrt {{x^2} - \sqrt {{x^2} - \dots} } } }}{x}$ is equal to-

$\lim _{n \rightarrow \infty} \frac{1}{n^3+1}+\frac{4}{n^3+1}+\frac{9}{n^3+1}+\ldots+\frac{n^2}{n^3+1} = $

$\lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$ is equal to

Find the limit: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}\right]$

The value of $\lim _{n}$ ${\rightarrow \infty}\left[\left(\frac{1}{2 \cdot 3}+\frac{1}{2^2 \cdot 3}\right)+\left(\frac{1}{2^2 \cdot 3^2}+\frac{1}{2^3 \cdot 3^2}\right)+\ldots+\left(\frac{1}{2^n \cdot 3^n}+\frac{1}{2^{n+1} \cdot 3^n}\right)\right]$ is