how to get the limit of $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)$ How to get the limit $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) = \frac{1}{2}$ ?
$\begin{align}
\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) &=
\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}}-\sqrt{n})(\sqrt{n+\sqrt{n}}+\sqrt{n})}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}+\sqrt{n}\sqrt{n+\sqrt{n}}-\sqrt{n}\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\  
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\frac{1}{\sqrt{n}}(\sqrt{n+\sqrt{n}}-n)}{\frac{1}{\sqrt{n}}(\sqrt{n+\sqrt{n}}+\sqrt{n})} \right) \\
&= \lim_{n \to \infty}\left( \frac{(\sqrt{\frac{n}{n}+\frac{\sqrt{n}}{n}}-\frac{n}{\sqrt{n}})}{(\sqrt{\frac{n}{n}+\frac{\sqrt{n}}{n}}+  \frac{\sqrt{n}}{\sqrt{n}})} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\frac{n}{\sqrt{n}}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\frac{\sqrt{n}\sqrt{n}}{\sqrt{n}}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\sqrt{n}}{2}\right) \\
&= \lim_{n \to \infty}\left(  \frac{1}{2} - \frac{\sqrt{n}}{2} \right) \\
\end{align}$
Which is wrong. 
Where could be my mistake?

To simplify the derivation you can use the binomial series and note that
$$\sqrt{n+\sqrt{n}}= \sqrt{n}\left( \sqrt{1+\frac1{\sqrt{n}}} \right)=\sqrt{n}\left( 1+\frac1{2\sqrt{n}}+o\left(\frac1{\sqrt{n}}\right) \right)=\sqrt{n}+\frac12+o(1)$$
thus
$$\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)=\sqrt{n}+\frac12+o(1)-\sqrt{n}=\frac12+o(1)\to\frac12$$
