The limit as $x \to \infty$ of $ \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}}$

Question

$$\lim_{x\to\infty} \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}}$$

I managed to find the limit of the above function, which according to my calculations is 1, but I don't know how to prove my answer is correct.

Answer 1

If you want a definition based proof ($M-\epsilon$) just note that for large $x$: $$ \left|\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}-1\right|=\\ \left|\frac{x+\sqrt{x}-1}{\sqrt{x+\sqrt{x}}\cdot \left(\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x+1}\right)\cdot\left(\sqrt{x+\sqrt{x}}+1\right)}\right|\leq\frac{2x}{x\sqrt{x}}=\frac{2}{\sqrt{x}}. $$

Answer 2

If you replace X by 1 / Y^2 and make a Taylor expansion around Y = 0, you should get it. I hope this is of some help to you.

Answer 3

Yes. Your answer is correct because $$ \lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x(1+\frac{\sqrt{x}}{x}}})}{\sqrt{x(1+\frac{1}{x})}}; $$

Because $\lim_{x\to\infty}\frac{1}{x}=0$, and $\lim_{x\to\infty}\frac{\sqrt{x}}{x}=0$ we have:

$$ \lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x(1+\frac{\sqrt{x}}{x}}})}{\sqrt{x(1+\frac{1}{x})}}=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x}}=\lim_{x\to\infty}\frac{\sqrt{x(1+\frac{\sqrt{x}}{x}})}{\sqrt{x}}=\lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x}}=1 $$

Answer 4

Since $$\sqrt{x}=_{\infty}o(x)$$ then simply we have $$ \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}}\sim_\infty \frac{\sqrt x}{\sqrt x}=1$$

Answer 5

For $x>0$ we have $$ \frac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt{x+\sqrt x}}=\frac{\sqrt x\cdot \sqrt{1+\frac{\sqrt{x+\sqrt x}}x}}{\sqrt x\cdot\sqrt{1+\frac1{\sqrt x}}} =\frac{\sqrt{1+\frac1{\sqrt x}\sqrt{1+\frac1{\sqrt x}}}}{\sqrt{1+\frac1{\sqrt x}}}.$$ Now as $x\to\infty$, all those $\frac1{\sqrt x}$ tend to $0$, hence all $\sqrt {1+\ldots}$ tend to $1$, hence so does the fraction.

Answer 6

That's obious: $$ \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}} =\sqrt{\frac{x+ \sqrt{ x+\sqrt {x}}}{x+1}} =\sqrt{\frac{1+\sqrt{1/x+1/\sqrt{x}}}{1+1/x}}.$$