# Limit of $\sqrt n (\sqrt{n+1}-\sqrt n)$ [closed]

How should I go about calculating $$\lim_{n \to \infty} \sqrt n (\sqrt{n+1}-\sqrt n)$$

I found empirically it approaches $$\frac 1 2$$ (and fast), but it's been decades since I last took a formal calculus class, and I don't know how to approach this.

• Multiply numerator and denominator by the conjugate $\sqrt{n+1}+\sqrt n$
– MasB
Commented Feb 25, 2021 at 15:50

Hint.

$$\sqrt n \left(\sqrt{n+1}-\sqrt n\right) = \frac{\sqrt n}{\sqrt{n+1} + \sqrt n} = \frac{1}{\sqrt{1 + \frac1n} + 1}$$

• Thank you, that's way more than a hint – it's precisely what I was looking for! Commented Feb 25, 2021 at 15:56
• Pretty nice hint +1 Commented Feb 3, 2023 at 17:44

Make use of $$\sqrt n (\sqrt {n+1}-\sqrt n)=\sqrt n (\sqrt {n+1}-\sqrt n)\frac{ \sqrt {n+1}+\sqrt n}{\sqrt {n+1}+\sqrt n}$$

This simplifies to $$\sqrt n\frac{ (\sqrt n+1)^{2}-\sqrt n^{2}}{\sqrt n+1)+\sqrt n}=\sqrt n\frac{n+1-n}{\sqrt{n+1}+\sqrt n}=\frac{\sqrt n}{\sqrt {n+1}+\sqrt n}=\frac{1}{\sqrt \frac{n+1}{n}+1}$$

This can be taken to the limit since $$\lim_{n\rightarrow \infty}\frac{n+1}{n}=1$$. There are several alternatives for this direct path. $$\lim_{n\rightarrow \infty}\frac{n+1}{n}=\lim_{n\rightarrow \infty}1+ \frac{1}{n}=1$$. This is calulating with limits. The denominator can not be developed for example into a Taylor series since one is not small.

So a substitution is well working: $$\frac{n+1}{n}=\left(\frac{m+1}{m}\right)^{2}$$. This takes some time to write down. Then the square root is gone and the limit again can be shown to be $$\frac{1}{2}$$. $$\lim_{m\rightarrow \infty}\frac{1}{2+\frac{1}{m}}=\frac{1}{2}$$