# 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 $\sqrt x$? – timon92 Dec 26 '17 at 20:28

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 – Paramanand Singh Dec 27 '17 at 5:20

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.

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. – egreg Feb 2 '18 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. – JJacquelin Feb 2 '18 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 – egreg Feb 2 '18 at 17:43
• @egreg Then the students need to have more redundant steps to understand the concept of limits :) – Mr Pie Feb 3 '18 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. – egreg Feb 3 '18 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. $$\Lambda = \lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x + \sqrt{x +\sqrt{x}}}} = \frac{\infty}{\infty} = 1$$ $$\therefore \Lambda = 1.\tag*{\bigcirc}$$