Prove that $\lim_{n \to \infty}\sqrt{n} \cdot \left(\sqrt{n+1} - \sqrt{n}\right)=\frac12$.

Question

This is a very unspecific and maybe stupid question, so I apologize for that. We recently had an exam that I failed, because I had pretty much no time to practice before that. Now I got to learn all that stuff that I should've known, yet there was one exercise where I have no clue how one would get a result.

Calculate $$\lim_{n \to \infty}\sqrt{n} \cdot (\sqrt{n+1} - \sqrt{n}).$$

The solution appears to be $\frac{1}{2}$, however I have exactly no clue on how to get to that.

If someone could throw a keyword at me, that would lead me on my path, it could already be very helpful.

Answer 1

Hint: $\displaystyle\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}=\frac1{\sqrt n}\times\frac1{\sqrt{1+\frac1n}+1}.$

Answer 2

Hint

$$\sqrt{n+1}-\sqrt n=\dfrac{n+1-n}{?}$$

Now set $1/n=h,h\to0^+$

Alternatively

$$\sqrt n(\sqrt{n+1}-\sqrt n)=\lim_{h\to0}\dfrac{\sqrt{1+h}-1}{h}$$

Set $\sqrt{1+h}=u$

Answer 3

$$\lim_{n \to \infty}\sqrt{n} \cdot (\sqrt{n+1} - \sqrt{n})= \lim_{n \to \infty}\sqrt{n} \cdot \frac{(\sqrt{n+1} - \sqrt{n})}{1}= \lim_{n \to \infty}\sqrt{n} \cdot \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{(\sqrt{n+1} + \sqrt{n})}= \lim_{n \to \infty}\sqrt{n}\cdot \frac{n+1-n}{\sqrt{n+1}+\sqrt n}= \lim_{n \to \infty}\sqrt{n}\cdot \frac1{\sqrt{n+1}+\sqrt n}= \lim_{n\to \infty}\frac{\sqrt{n}}{\sqrt n}\cdot\frac1{\sqrt{1+\frac1n}+1}=\lim_{n \to \infty} \frac1{\sqrt{1+\frac1n}+1}= \frac1{2}$$

Answer 4

$$\lim_{h\to0^+}\frac{\sqrt{\dfrac 1h+1}-\sqrt{\dfrac 1h}}{\sqrt h}=\lim_{h\to0}\frac{\sqrt{h+1}-1}h=\left.(\sqrt{x+1})'\right|_{x=0}=\frac12.$$