Answer
\[2[\sqrt 2-1]\]
Work Step by Step
We have to find :-
\[\lim_{n\rightarrow \infty}\left[\frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+...+\frac{1}{\sqrt{n}\sqrt{n+n}}\right]\]
\[=\lim_{n\rightarrow \infty}\left[\frac{1}{n\sqrt{1+\frac{1}{n}}}+\frac{1}{n\sqrt{1+\frac{2}{n}}}+...+\frac{1}{n\sqrt{1+\frac{n}{n}}}\right]\]
\[=\lim_{n\rightarrow \infty}\frac{1}{n}\left[\frac{1}{\sqrt{1+\frac{1}{n}}}+\frac{1}{\sqrt{1+\frac{2}{n}}}+...+\frac{1}{\sqrt{1+\frac{n}{n}}}\right]\]
\[=\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n}}}\;\;\;...(1)\]
We will use the formula \[\lim_{n\rightarrow \infty}\frac{1}{n} \sum_{k=f(n)}^{g(n)}F\left(\frac{k}{n}\right)=\int_{\lim_{n\rightarrow \infty}\frac{f(n)}{n}}^{
\lim_{n\rightarrow \infty}\frac{g(n)}{n}}F(x)\:dx\;\;\;...(2)\]
Using (2) in (1)
\[\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n}}}=\int_{\lim_{n\rightarrow \infty}\frac{1}{n}}^{\lim_{n\rightarrow \infty}\frac{n}{n}}\frac{1}{\sqrt{1+x}}dx\]
\[\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n}}}=\int_{0}^{1}\frac{1}{\sqrt{1+x}}dx\]
\[\Rightarrow \lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n}}}=2\left[\sqrt{1+x}\right]_{0}^{1}\]
\[\Rightarrow \lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n}}}=2\left[\sqrt{2}-1\right]\]
Hence \[\lim_{n\rightarrow \infty}\left[\frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+...+\frac{1}{\sqrt{n}\sqrt{n+n}}\right]=2\left[\sqrt{2}-1\right]\]
