how to get the limit of $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)$

Question

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?

Answer 1

$$\begin{align} \sqrt{n+\sqrt{n}}-\sqrt{n} &= \frac{(\color{red}{\sqrt{n+\sqrt{n}}}-\color{blue}{\sqrt{n}})(\color{red}{\sqrt{n+\sqrt{n}}}+\color{blue}{\sqrt{n}})}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\ &= \frac{(\color{red}{\sqrt{n+\sqrt{n}}})^2-(\color{blue}{\sqrt{n}})^2}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\ &= \frac{(n+\sqrt{n})-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\ &= \frac{\sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\ &= \frac{1}{\sqrt{1+1/\sqrt{n}}+1} \\ \end{align}$$

and $1/\sqrt n \to 0$ hence $$\lim_{n\to\infty}\sqrt{n+\sqrt{n}}-\sqrt{n} = \lim_{n\to\infty}\frac{1}{\sqrt{1+1/\sqrt{n}}+1} = \frac 1{\sqrt{1+0}+1} =\frac 12$$

Answer 2

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

$$\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)=\sqrt{n}+\frac12+o(1)-\sqrt{n}=\frac12+o(1)\to\frac12$$

Answer 3

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
$$\lim_{n \to \infty}\left( \frac{{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) $$

Answer 4

In a simpler way:

Let $n=m^2$.

$$\lim_{m\to\infty}\sqrt{m^2+m}-m=\lim_{m\to\infty}\frac m{\sqrt{m^2+m}+m}=\lim_{m\to\infty}\frac 1{\sqrt{1+\dfrac1m}+1}.$$