limit $\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 2+\sqrt4}+\frac{1}{\sqrt4+\sqrt6}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right)$

Note that

$$\frac{1}{\sqrt{k}+\sqrt{k+2}}=\frac{\sqrt{k+2}-\sqrt{k}}{2}$$

So we have

\begin{align} &\;\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 2+\sqrt4}+\frac{1}{\sqrt4+\sqrt6}+\cdots+\frac{1}{\sqrt2n+\sqrt{2n+2}}\right)\\ =&\;\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{\sqrt 4-\sqrt2}{2}+\frac{\sqrt6-\sqrt4}{2}+\cdots+\frac{\sqrt{2n+2}-\sqrt{2n}}{2}\right)\\ =&\;\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{\sqrt{2n+2}-\sqrt{2}}{2}\right)\\ =&\;\lim_{n\to\infty}\left(\frac{\sqrt{2+\frac{2}{n}}-\sqrt{\frac{2}{n}}}{2}\right)\\ =&\;\frac{\sqrt{2}}{2} \end{align}

HINT:$$\frac{1}{\sqrt{2n}+\sqrt{2n+2}}=\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\cdot\frac{\sqrt{2n+2}-\sqrt{2n}}{\sqrt{2n+2}-\sqrt{2n}}=\frac{\sqrt{2n+2}-\sqrt{2n}}{2}$$

Hint: write $$\frac{1}{\sqrt{2n}+\sqrt{2n+2}}=\frac{\sqrt{2n}-\sqrt{2n+2}}{-2}$$ etc