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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

Divide by $\sqrt{x}$ to get

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

Answer 4

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*}

Answer 5

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}}}= $

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

Answer 6

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

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