# Nested radicals; $\lim\limits_{x \to \infty}{}\sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x}$

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.

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

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 $y \to 0$.
With the substitution $x=1/t^2$, the limit is the same as $$\lim_{t\to0^+} \left( \sqrt{\frac{1}{t^2}+\sqrt{\frac{1}{t^2}+\sqrt{\frac{1}{t^2}}}}-\sqrt{\frac{1}{t^2}} \right)= \lim_{t\to0^+}\frac{\sqrt{1+t\sqrt{1+t}}-1}{t}= \lim_{t\to0^+}\frac{\frac{1}{2}t\sqrt{1+t}+o(t\sqrt{1+t}\,)}{t}$$