I tried conjugating and it didn't lead me anywhere please help guys.


  • $\begingroup$ I edited the mathematical part of your question using MathJax. Please, use it in your future questions. However, it's unclear if $x\to \infty$ or $x\to 0$ from your description. Please, add this detail, too. $\endgroup$ – TZakrevskiy Jan 6 '14 at 14:39
  • $\begingroup$ Do you mean $\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt x}}}-\sqrt x$? $\endgroup$ – Yiorgos S. Smyrlis Jan 6 '14 at 14:41
  • $\begingroup$ yes lim x-infinite $\endgroup$ – niako1 Jan 6 '14 at 14:44
  • $\begingroup$ I believe this is a duplicate: I'm pretty sure I've seen it a few weeks ago. $\endgroup$ – egreg Jan 6 '14 at 14:44

You can get the following :

$$\begin{align}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt x}}}-\sqrt x&=\frac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt{x+\sqrt{x+\sqrt{x+\sqrt x}}}+\sqrt x}\\&=\frac{\sqrt{1+\left(\sqrt{x+\sqrt x}\right)/x}}{\sqrt{1+\left(\sqrt{x+\sqrt{x+\sqrt x}}\right)/x}+1}\end{align}$$

Now divide both the numerator and the dinominator by $\sqrt x$.

  • $\begingroup$ if I divide by sqrtx It will by infinite/ninfinite $\endgroup$ – niako1 Jan 6 '14 at 14:52
  • $\begingroup$ @user119400: I added a bit. $\endgroup$ – mathlove Jan 6 '14 at 14:55
  • $\begingroup$ and afterthat? it keep getting infinite/infinite $\endgroup$ – niako1 Jan 6 '14 at 15:05
  • $\begingroup$ No. $(\sqrt{x+\sqrt x})/x\to 0$, $(\sqrt{x+\sqrt{x+\sqrt x}})/x\to 0$ when $x\to\infty$. So, the answer will be $1/(1+1)$. $\endgroup$ – mathlove Jan 6 '14 at 15:06
  • $\begingroup$ Why lim sqrt(x+sqrtx))=0 ? $\endgroup$ – niako1 Jan 6 '14 at 15:12

I think the best way to do it is to note that

$$\sqrt{x + \sqrt{x}} \sim \sqrt{x}$$

So you can write that $\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}$ = $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x}$

(We can't simplify any further cause that would bring us to $\sqrt{x} - \sqrt{x}$ which doesn't make sense.)

Then write as

$$\sqrt{x} \cdot (\sqrt{1 + \frac{1}{\sqrt{x}}} - 1) \sim \sqrt{x} \cdot (1 + \frac{1}{2\sqrt{x}} - 1) = \frac{1}{2}$$

  • $\begingroup$ According to my math teacher, you can't do that simplification. You are no allowed to, a priori, "know" that the limit of $\sqrt{x+\sqrt{x}}$ is the same as $\sqrt{x}$ as $x \to \infty$. $\endgroup$ – Björn Lindqvist Feb 26 '17 at 19:58
  • $\begingroup$ @BjörnLindqvist What do you mean by a priori? I can just compute the limit $$\lim_{x \to \infty} \frac{\sqrt{x + \sqrt x}}{\sqrt x} = 1$$ which should be easy enough as a first step and then continue with the asymptotic argument. But in general if you already know this holds (because you've done it already other times) then it is very easy to see that the original limit simplifies to $\lim_{x \to \infty} \sqrt{x + \sqrt x} - \sqrt x $ and proceed :) Whether it is necessary in a solution to an exercise to prove that $\sqrt{x + \sqrt x} \sim \sqrt x$ or not it's up to the teacher $\endgroup$ – Ant Feb 26 '17 at 20:07
  • $\begingroup$ What I mean is that in your solution, you didn't prove that $\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}$ = $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x}$ therefore it must not be accepted as a fact. And (this is what I think is my teachers point) showing that $\lim_{x\to\infty}{\sqrt{x}} = \infty$ and $\lim_{x\to\infty}{\sqrt{x+\sqrt{x}}} = \infty$ doesn't prove the above mentioned equality. $\endgroup$ – Björn Lindqvist Feb 26 '17 at 21:59
  • $\begingroup$ @BjörnLindqvist Oh absolutely!But when I write that $\sqrt{x + \sqrt x} \sim \sqrt x$ I don't mean that both those expressions tend to the same limit;it is a much stronger condition,which implies certain things about calculating limits by substituting expression which are in a $\sim$ relation.In particular you can (usually) substitute functions which are asymptotic to each other (en.wikipedia.org/wiki/Asymptotic_analysis).I say usually because it don't always hold,but you can look up when the substitution is valid for calculating limits,in which case it helps the computation enormously $\endgroup$ – Ant Feb 26 '17 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.