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?
 A: $$\begin{align}
\sqrt{n+\sqrt{n}}-\sqrt{n} &=
\frac{(\color{red}{\sqrt{n+\sqrt{n}}}-\color{blue}{\sqrt{n}})(\color{red}{\sqrt{n+\sqrt{n}}}+\color{blue}{\sqrt{n}})}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\
&= \frac{(\color{red}{\sqrt{n+\sqrt{n}}})^2-(\color{blue}{\sqrt{n}})^2}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\
&= \frac{(n+\sqrt{n})-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\
&= \frac{\sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\
&= \frac{1}{\sqrt{1+1/\sqrt{n}}+1} \\
\end{align}$$
and $1/\sqrt n \to 0$ hence
$$\lim_{n\to\infty}\sqrt{n+\sqrt{n}}-\sqrt{n} = \lim_{n\to\infty}\frac{1}{\sqrt{1+1/\sqrt{n}}+1} = \frac 1{\sqrt{1+0}+1} =\frac 12$$
A: In a simpler way:
Let $n=m^2$.
$$\lim_{m\to\infty}\sqrt{m^2+m}-m=\lim_{m\to\infty}\frac m{\sqrt{m^2+m}+m}=\lim_{m\to\infty}\frac 1{\sqrt{1+\dfrac1m}+1}.$$
A: 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
$$\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)=\sqrt{n}+\frac12+o(1)-\sqrt{n}=\frac12+o(1)\to\frac12$$
A: The third line which is 
$$\lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) $$
should be
$$\lim_{n \to \infty}\left( \frac{{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) $$
