Evaluate the following limit: Find
$$\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left[\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right]$$
MY TRY:
$$
\begin{align}
\lim_{n \to \infty} &\frac{1}{\sqrt{n}} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}} 
+ \frac{1}{\sqrt{4}+\sqrt{6}} 
+ \cdots + \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr] \\
&= \lim_{n \to \infty} \frac{\sqrt{n}}{n} \biggl[ 
\frac{1}{\sqrt{2}+\sqrt{4}} 
+ \frac{1}{\sqrt{4}+\sqrt{6}} + \cdots 
+ \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr]
\end{align}
$$
Now using Cauchy first thm of limits
$$
a_n = \frac{\sqrt{n}}{\sqrt{2n}+\sqrt{2n+1}}
$$
The answer should be $\frac{1}{2\sqrt{2}}$.
But the answer is $1/\sqrt{2}$.

This is a classic partial sums of integrals problem. Your expression is:
$$\frac{1}{\sqrt{n}}\sum\limits_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$$
write as 
$$\frac{1}{n}\sum\limits_{k=1}^n \frac{1}{\sqrt{2\frac{k}{n}}+\sqrt{2\frac{k}{n}+2}}$$
then the limit is the same as the integral
$$\int_0^1\frac{1}{\sqrt{2x}+\sqrt{2x+2}}dx=
\int_0^1\frac{\sqrt{2x}-\sqrt{2x+2}}{-2}dx
$$
