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?

• In your second line, something should have been squared (i.e., $(a+b)(a-b) = a^2 - b^2$). You forgot to remove the first square root. Also, there was no need to show the middle terms, since you can just use the standard difference-of-two-squares result. – quasi Jan 29 '18 at 11:14
• This doesn't answer the question, but you can use Taylor series. If we let the limit go to $n+1$, then we can use the Taylor series of $\sqrt{n+1}$ which is $1+\frac{n}{2}-\frac{n^2}{8} \cdots$, and then 'plug' that in to $\sqrt{n+1+\sqrt{n+1}}$. – Toby Mak Jan 29 '18 at 11:19
• Indeed, the second line should be $$\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}})^2+\sqrt{n}\sqrt{n+\sqrt{n}}-\sqrt{n}\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}}\right).$$ – Professor Vector Jan 29 '18 at 11:20
• @jublikon let me know if the alternative solution is useful for you, otherwise I'will delete it. – gimusi Jan 29 '18 at 11:27
• @gimusi unfortunately I am not familiar with your technique. But I think it could be useful for other readers – jublikon Jan 29 '18 at 11:52

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

In a simpler way:

Let $n=m^2$.

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

• very nice trick! – gimusi Jan 29 '18 at 11:31

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

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

• Please answer the question – the OP wants to find where he went wrong in his proof. – Toby Mak Jan 29 '18 at 11:24
• @TobyMak the OP is "How to get the limit?". The error has been already pointed out in the comments thus I think it can be useful to give an alternative and effective way to solve it. – gimusi Jan 29 '18 at 11:26

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