$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \dots + \frac{1}{{2n}}} \right] = $

$\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}}\right]=$

$\lim _{n \rightarrow \infty}\left[\frac{\sqrt{n^2-1^2}}{n^2}+\frac{\sqrt{n^2-2^2}}{n^2}+\frac{\sqrt{n^2-3^2}}{n^2}+\ldots+\frac{\sqrt{n^2-n^2}}{n^2}\right]=$

$\mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 0}^n {\frac{n}{{{{\left( {2r + n} \right)}^2}}}} $ ની કિંમત શોધો.

$\lim _{n \rightarrow \infty}\left(\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots+\frac{1}{2 n}\right)$ ની કિંમત શોધો :-

$\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right) \ldots \left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=$