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I have the following problem: $$ \lim_{x \to \infty}{\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}=? $$

The answer in the answer section is $\frac{1}{2}$.

What I've tried:

$$ \lim_{x \to \infty}\sqrt{x+\sqrt{x+\sqrt{x}}}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}\frac{(x-\sqrt{x+\sqrt{x}})}{(x-\sqrt{x+\sqrt{x}})} $$

I tried a couple of other simplifying approaches but unfortunately nothing got me near a limit of $\frac{1}{2}$, which makes me think that it's some other approach I'm unaware of.

Edit: Sorry for the error in the problem. It's how it was in the book and not my fault in rewriting it.

Thanks in advance!

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    $\begingroup$ There must be something wrong with the question, since this is obviously unbounded. What is the real sequence? Is it $\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\dotso}}}}}}$ $\endgroup$
    – Cornman
    Nov 27, 2017 at 18:25
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    $\begingroup$ The function is greater than $\sqrt{x}$, so the limit is $\infty$. Check the source to make sure you've stated the problem correctly. $\endgroup$
    – quasi
    Nov 27, 2017 at 18:26
  • $\begingroup$ How is there a limit? It looks like the function increases without bound. Did we forget a term? $\endgroup$
    – Oscar Lanzi
    Nov 27, 2017 at 18:26
  • $\begingroup$ How is there a limit? It looks like the function increases without bound. Did we forget a term? $\endgroup$
    – Oscar Lanzi
    Nov 27, 2017 at 18:26
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    $\begingroup$ Or is it just $$\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt x}}-\sqrt x=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt x}}{\sqrt{x+\sqrt{x+\sqrt x}}+\sqrt x}=\frac12?$$ $\endgroup$
    – Lutz Lehmann
    Nov 27, 2017 at 19:09

2 Answers 2

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As $x \to \infty$, $$ \sqrt{x+\sqrt{x}} = \sqrt{x}\;\big({1+x^{-1/2}}\big)^{1/2} = x^{1/2}\big(1+O(x^{-1/2})\big) = x^{1/2}+ O(1) $$ and then $$ \sqrt{x+\sqrt{x+\sqrt{x}}} = \sqrt{x}\sqrt{1+\frac{x^{1/2} + O(1)}{x}} = x^{1/2}\left(1+x^{-1/2}+O(x^{-1})\right)^{1/2} \\ x^{1/2}\left(1+\frac{1}{2}\;x^{-1/2}+O(x^{-1})\right) = x^{1/2}+\frac{1}{2} + O(x^{-1/2}) $$ so that $$ \sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x} = \frac{1}{2} + O(x^{-1/2}) $$ and we get limit $\frac{1}{2}$.

We used, twice, $\sqrt{1+y} = 1 + \frac{1}{2}y + O(y^2)$ as $y \to 0$.

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With the substitution $x=1/t^2$, the limit is the same as $$ \lim_{t\to0^+} \left( \sqrt{\frac{1}{t^2}+\sqrt{\frac{1}{t^2}+\sqrt{\frac{1}{t^2}}}}-\sqrt{\frac{1}{t^2}} \right)= \lim_{t\to0^+}\frac{\sqrt{1+t\sqrt{1+t}}-1}{t}= \lim_{t\to0^+}\frac{\frac{1}{2}t\sqrt{1+t}+o(t\sqrt{1+t}\,)}{t} $$

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