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


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.

  • 2
    Can you add your working out/thinking to the question? Try and extend your knowledge with 2 to what could happen with 3. You are adding them so the behavior is very similar. – MRobinson Oct 11 at 14:39
  • If $x$ is close to zero we have $\sqrt{1+x}=1+\frac{x}{2}+O(x^2)$, hence $$\sqrt{4n+1}-\sqrt{n}-\sqrt{n+1} = \sqrt{n}\left[\frac{1}{4n}-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right]$$ clearly converges to $\ldots$ when $n\to +\infty$. – Jack D'Aurizio Oct 11 at 17:33
up vote 11 down vote accepted

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

Set $1/n=h^2$


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

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

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

So, the original limit is $0$.

Use the "conjugate" trick twice.

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


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

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

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