# Question #1f1f7

Sep 30, 2017

${\lim}_{x \rightarrow \infty} \frac{\sqrt{x + \sqrt{x + \sqrt{x}}}}{\sqrt{x + 1}} = 1$

#### Explanation:

Let's try to factor within the square roots:

${\lim}_{x \rightarrow \infty} \frac{\sqrt{x + \sqrt{x + \sqrt{x}}}}{\sqrt{x + 1}}$

$= {\lim}_{x \rightarrow \infty} \frac{\sqrt{x + \sqrt{x \left(1 + \frac{1}{\sqrt{x}}\right)}}}{\sqrt{x \left(1 + \frac{1}{x}\right)}}$

Pull out what we've just factored from their respective square roots:

$= {\lim}_{x \rightarrow \infty} \frac{\sqrt{x + \sqrt{x} \sqrt{1 + \frac{1}{\sqrt{x}}}}}{\sqrt{x} \sqrt{1 + \frac{1}{x}}}$

Factor from the numerator again:

$= {\lim}_{x \rightarrow \infty} \frac{\sqrt{x \left(1 + \frac{1}{\sqrt{x}} \sqrt{1 + \frac{1}{\sqrt{x}}}\right)}}{\sqrt{x} \sqrt{1 + \frac{1}{x}}}$

And pull this from the numerator:

$= {\lim}_{x \rightarrow \infty} \frac{\sqrt{x} \sqrt{1 + \frac{1}{\sqrt{x}} \sqrt{1 + \frac{1}{\sqrt{x}}}}}{\sqrt{x} \sqrt{1 + \frac{1}{x}}}$

Which cancels:

$= {\lim}_{x \rightarrow \infty} \frac{\sqrt{1 + \frac{1}{\sqrt{x}} \sqrt{1 + \frac{1}{\sqrt{x}}}}}{\sqrt{1 + \frac{1}{x}}}$

Note that both $\frac{1}{x}$ and $\frac{1}{\sqrt{x}}$ approach $0$ as $x$ approaches infinity.

$= \frac{\sqrt{1 + 0 \sqrt{1 + 0}}}{\sqrt{1 + 0}}$

$= \frac{\sqrt{1}}{\sqrt{1}}$

$= 1$