$\mathop {\lim }\limits_{n \to \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}} = $

$\mathop {\lim }\limits_{n \to \infty } {\left\{ {\left( {1 + \frac{{{1^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{2^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{3^2}}}{{{n^2}}}} \right) \dots \left( {1 + \frac{{{{(n - 1)}^2}}}{{{n^2}}}} \right)} \right\}^{1/n}}$ ની કિંમત શોધો:

ધારો કે $\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \left( \frac{n}{\sqrt{n^4+r^4}} - \frac{2 n r^2}{(n^2+r^2) \sqrt{n^4+r^4}} \right) = \frac{\pi}{k}.$ પ્રતિ-ત્રિકોણમિતીય વિધેયોના મુખ્ય મૂલ્યોનો ઉપયોગ કરીને,તો $k^2$ ની કિંમત શોધો:

$\lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{4 n^2-1^2}}+\frac{1}{\sqrt{4 n^2-2^2}}+\frac{1}{\sqrt{4 n^2-3^2}}+\dots+\frac{1}{\sqrt{4 n^2-n^2}}\right\}=$

$\lim _{n \rightarrow \infty}\left(\frac{n^{2}}{\left(n^{2}+1\right)(n+1)}+\frac{n^{2}}{\left(n^{2}+4\right)(n+2)}+\frac{n^{2}}{\left(n^{2}+9\right)(n+3)}+\ldots+\frac{n^{2}}{\left(n^{2}+n^{2}\right)(n+n)}\right)$ ની કિંમત શોધો.

જો $\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=k$ હોય,તો $\log k=$