# Find: $\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$

Find: $\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$

Question from a book on preparation for math contests. All the tricks I know to solve this limit are not working. Wolfram Alpha struggled to find $1$ as the solution, but the solution process presented is not understandable. The answer is $1$.

Hints and solutions are appreciated. Sorry if this is a duplicate.

• Have you tried to divide both nominator and denominator by $\sqrt x$? Dec 26, 2017 at 20:28

Note that $$\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}} = \frac{1}{\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^3}}}}}$$

You can also look at it as $$\sqrt{\frac{x}{x+\sqrt{x+\sqrt{x}}}}$$ In case dividing by $\sqrt{x}$ bothers you.

A fun overkill: it is well known (at least among Ramanujan supporters) that for any $x>1$ we have $$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \tfrac{1}{2}+\sqrt{x+\tfrac{1}{4}}$$ hence $\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}$ is bounded between $1$ and $\frac{\sqrt{x}}{\sqrt{x+\frac{1}{4}}+\frac{1}{2}}$, whose limit as $x\to +\infty$ is also $1$.
The claim hence follows by squeezing.

• Nice to see Ramanujan's name here. I am one of his supporters :) +1 Dec 27, 2017 at 5:20

Divide by $\sqrt{x}$ to get

$$\lim_{x \to \infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \sqrt{\frac 1x + \sqrt{\frac{1}{x^3}}}}} = 1$$

If you factor out a $\sqrt{x}$ term from the denominator one has \begin{align*} \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}}} &= \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x} \sqrt{1 + \frac{1}{x} \sqrt{x + \sqrt{x}}}}\\ &= \lim_{x \to \infty} \frac{1}{\sqrt{1 + \sqrt{\frac{1}{x} + \frac{1}{x^{3/2}}}}}\\ &= 1. \end{align*}

$$\text{Let}\quad x=\frac{1}{\epsilon^2} \quad\implies\quad \frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=\frac{1}{1+\epsilon\:\sqrt{1+\epsilon}}\qquad\qquad \epsilon\neq 0$$

$$\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}} = \lim_{\epsilon\to 0}\frac{1}{1+\epsilon\:\sqrt{1+\epsilon}} = \lim_{\epsilon\to 0}\frac{1}{1+\epsilon\sqrt{1}} = \lim_{\epsilon\to 0}\frac{1}{1+\epsilon} =1$$

• Why not directly $\lim\limits_{\epsilon\to 0}\frac{1}{1+\epsilon\,\sqrt{1+\epsilon}} =1$? There are two redundant steps. Feb 2, 2018 at 16:21
• Just a teaching care. For you it is obvious that $\epsilon\sqrt{1+\epsilon}\to 0$. But this in not so obvious for some students who need more intermediate steps. Feb 2, 2018 at 16:53
• There is no intermediate step here! This limit can and should be computed by direct substitution. Would you do $\lim\limits_{x\to0}\frac{x-\sin x}{x^3}=\lim\limits_{x\to0}\frac{x-0}{x^3}$? I hope not, but your steps could give this idea to the students Feb 2, 2018 at 17:43
• @egreg Then the students need to have more redundant steps to understand the concept of limits :) Feb 3, 2018 at 1:14
• @user477343 The problem is that in general one cannot substitute only parts of an expression with its limit. In my opinion, these particular intermediate steps cal lead to misunderstandings. Feb 3, 2018 at 10:13

Let $y=√x$. $\lim x \rightarrow \infty =\lim y \rightarrow \infty.$

Numerator: $y$

Denominator:

$\sqrt {y^2 +\sqrt{y^2+y}}= \sqrt{y^2+y\sqrt{1+1/y}}=$

$y\sqrt{1+(1/y)\sqrt{1+1/y}}.$

$\lim_{y \rightarrow \infty} \dfrac{y}{y \sqrt{1+(1/y) \sqrt{1+1/y}}}=$

$\lim_{y \rightarrow \infty} \dfrac{1}{\sqrt{1+(1/y)\sqrt{1+1/y}}} =1.$

Let $\Lambda =$ the limit we need to find. Then, $\ \Box\ \Lambda = 1$.

Proof: We will begin our proof using the following Lemma. $$\forall a, b\in\mathbb{R}, \ \sqrt{a + \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} + \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}.\tag1$$ Substitute $a = x$ and $b = x + \sqrt{x}$ into the Lemma. $$(1) = \sqrt{\frac{x + \sqrt{x^2 - x + \sqrt{x}}}{2}} + \sqrt{\frac{x - \sqrt{x^2 - x + \sqrt{x}}}{2}}.$$ Find the limit of the fractions under each root.$$\lim_{x\to\infty}\frac{x \pm \sqrt{x^2 - x + \sqrt{x}}}{2} = \frac 12\lim_{x\to\infty}\bigg(x \pm \sqrt{x^2 - x + \sqrt{x}}\bigg) = \frac12\cdot\infty = \infty$$ $$\therefore \lim_{x\to\infty}\sqrt{x + \sqrt{x + \sqrt{x}}} = \sqrt{\infty} + \sqrt{\infty} = \infty + \infty = \infty.$$ And, $\because \lim_{x\to\infty}\sqrt{x} = \infty$ then we finally have as desired. $$\Lambda = \lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x + \sqrt{x +\sqrt{x}}}} = \frac{\infty}{\infty} = 1$$ $$\therefore \Lambda = 1.\tag*{\bigcirc}$$

• @downvoter care to explain why I was downvoted? Is there a way I could improve my answer? :) Jul 28, 2019 at 23:33