How to calculate the limit $\lim_{n\rightarrow\infty}\left(\sqrt{4n+1}-\sqrt{n}-\sqrt{n+1}\right)$?

Question

I have a task for school to calculate this limit at infinity, I have tried three times but I failed every time.

$$\lim_{n\rightarrow\infty}\left(\sqrt{4n+1}-\sqrt{n}-\sqrt{n+1}\right)$$

I know what to do when there are two square roots but when there's three I don't know how to proceed. Can anyone help me? Thanks.

Answer 1

Hint: Since you know what to do when there are only two square roots, use the fact that$$(\forall n\in\mathbb{N}):\sqrt{4n+1}-\sqrt n-\sqrt{n+1}=\left(\sqrt{n+\frac14}-\sqrt n\right)+\left(\sqrt{n+\frac14}-\sqrt{n+1}\right).$$

Answer 2

Write $$\frac{((\sqrt{4n+1}-\sqrt{n+1})-\sqrt{n})((\sqrt{4n+1}-\sqrt{n+1})+\sqrt{n})}{\sqrt{4n+1}-\sqrt{n+1}+\sqrt{n}}$$

Answer 3

$\sqrt{4n+1}-(\sqrt{n}+\sqrt{n+1})<\sqrt{4n+1}-2\sqrt{n}=\sqrt{4n+1}-\sqrt{4n} <\frac{1}{\sqrt{4n+1}+\sqrt{4n}}\to0$

So, the original limit is $0$.

Answer 4

Set $1/n=h^2$

$$\lim_{h\to0^+}\dfrac{\sqrt{4+h^2}-1-\sqrt{1+h^2}}h$$

$$=\lim_{...}\dfrac{\sqrt{4+h^2}-2}h-\lim_{...}\dfrac{\sqrt{1+h^2}-1}h=0-0$$ on rationalization of the numerators

Answer 5

Use the "conjugate" trick twice.

$$\sqrt{4n+1}-\sqrt n-\sqrt{n+1}=\frac{(\sqrt{4n+1}-\sqrt n)^2-(n+1)}{(\sqrt{4n+1}-\sqrt n)+\sqrt{n+1}}$$

and

$$4n+1+n-2\sqrt{4n+1}\sqrt n-(n+1)=4n-2\sqrt{4n+1}\sqrt n=\frac{16n^2-4(4n+1)n}{4n+2\sqrt{4n+1}\sqrt n}.$$

This last expression tends to a finite value (namely $-\frac12$), so that the initial expression is asymptotic to $-\frac14n^{-1/2}$.