The limit as $x \to \infty$ of $ \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}}$ $$\lim_{x\to\infty} \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}}$$
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.
 A: Since
$$\sqrt{x}=_{\infty}o(x)$$
then simply we have
$$ \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}}\sim_\infty \frac{\sqrt x}{\sqrt x}=1$$
A: 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.
A: If you want a definition based proof ($M-\epsilon$) just note that for large $x$:
$$
\left|\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}-1\right|=\\
\left|\frac{x+\sqrt{x}-1}{\sqrt{x+\sqrt{x}}\cdot \left(\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x+1}\right)\cdot\left(\sqrt{x+\sqrt{x}}+1\right)}\right|\leq\frac{2x}{x\sqrt{x}}=\frac{2}{\sqrt{x}}.
$$
A: 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.
A: That's obious:
$$ \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}}
=\sqrt{\frac{x+ \sqrt{ x+\sqrt  {x}}}{x+1}}
=\sqrt{\frac{1+\sqrt{1/x+1/\sqrt{x}}}{1+1/x}}.$$
A: 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:
$$
\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x(1+\frac{\sqrt{x}}{x}}})}{\sqrt{x(1+\frac{1}{x})}}=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x}}=\lim_{x\to\infty}\frac{\sqrt{x(1+\frac{\sqrt{x}}{x}})}{\sqrt{x}}=\lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x}}=1
$$
