જો $f(n) = \frac{1}{n} [(n+1)(n+2)(n+3) \ldots (2n)]^{\frac{1}{n}}$ હોય, તો $\lim_{n \rightarrow \infty} f(n) =$

જો $f: R \rightarrow R$ એ $f(x)=x+1$ દ્વારા વ્યાખ્યાયિત હોય,તો $\lim _{n \rightarrow \infty} \frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+\ldots+f\left(\frac{5(n-1)}{n}\right)\right]$ નું મૂલ્ય શોધો.

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

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{\left( {n + 2} \right)}^{1/3}}}}{{{n^{4/3}}}} + \dots + \frac{{{{\left( {2n} \right)}^{1/3}}}}{{{n^{4/3}}}}} \right)$ ની કિંમત શોધો.

$\lim _{n \rightarrow \infty}\left[\frac{n}{(n+1) \sqrt{2n+1}}+\frac{n}{(n+2) \sqrt{2(2n+2)}}+\frac{n}{(n+3) \sqrt{3(2n+3)}}+\ldots n \text{ પદો}\right]=\int_0^1 f(x) d x$,તો $f(x)=$

$\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]=$