How to solve $\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$ Solved I stuck in this limit, I tried to solve it and gets 1/2 as result. Yet, I was wrong because I forgot a square. Please need help!
$\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$
Note: it's $+\infty$ 
Thanks in advance 
Update: I actually solved it, and this is the way that I wanted to:
$\lim \:_{x\to \:\infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)\:=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x+\sqrt{1+\sqrt{x}}}+\sqrt{x}}\right)=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x^2\left(1+\frac{\sqrt{1+\sqrt{x}}}{x}\right)+\sqrt{x}}}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x\left(\frac{1}{x}+\frac{1}{\sqrt{x}}\right)}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x}\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{x}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1}\right)=\frac{0}{2}=0$
Thanks for your two answers guys. I really appreciate it ! This community is awesome !
 A: $$\begin{align}\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x&=\frac{\left(\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x\right)\left(\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x\right)}{\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x}\\
&\le \frac{\left(\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x\right)\left(\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x\right)}{1+\sqrt x}\\
&=\frac{\sqrt{1+\sqrt x}}{1+\sqrt x}\\
&=\frac1{\sqrt{1+\sqrt x}}\\
&\to 0\end{align}$$
A: If ${\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x} = 0}$ and you are taking a limit of the form
$$\lim_{x \rightarrow \infty} \sqrt{x + f(x)} - \sqrt{x}$$
Then you can proceed by writing the limit as
$$\lim_{x \rightarrow \infty} \sqrt{x} \bigg(\sqrt{1 + {f(x) \over x}} - 1\bigg)$$
Since $\sqrt{1 + \epsilon} = 1 + {1 \over 2} \epsilon + O(\epsilon^2)$, the limit will be the same as 
$$\lim_{x \rightarrow \infty} \frac{f(x)}{2\sqrt{x}}$$
In the case at hand, $f(x) = \sqrt{1 + \sqrt{x}}$, which is bounded by
for example $\sqrt{\sqrt{x} + \sqrt{x}} = \sqrt 2 x^{1 \over 4}$. Hence the limit will be zero. 
A: For clarity, I will use $y=\sqrt{x}$ The limit becomes
$$
\lim_{y\to \infty}( \sqrt{y^2+\sqrt{1+y}}-y)= \lim_{y\to \infty}( \sqrt{y^2+\sqrt{1+y}}-y)\frac{\sqrt{y^2+\sqrt{1+y}}+y}{\sqrt{y^2+\sqrt{1+y}}+y}
=\lim_{y\to \infty} \frac{\sqrt{1+y}}{\sqrt{y^2+\sqrt{1+y}}+y} \\
=\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{\sqrt{y^2+y^2\sqrt{1/y^4+1/y^3}}+y}\\
 =\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{\sqrt{y^2(1+\sqrt{1/y^4+1/y^3})}+y}\\
=\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{|y|\sqrt{1+\sqrt{1/y^4+1/y^3}}+y}\\
=\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{y\sqrt{1+\sqrt{1/y^4+1/y^3}}+y}\\
=\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{y\left(\sqrt{1+\sqrt{1/y^4+1/y^3}}+1\right)}\\
=\lim_{y\to \infty}\frac{\sqrt{1/y+1}}{\sqrt{y}\left(\sqrt{1+\sqrt{1/y^4+1/y^3}}+1\right)}\\
\to \frac{1}{\infty \cdot 2}=0
$$
A: If we have $f(x)/x^{1/2} \to 0,$ where $f(x) > 0,$ then the MVT gives
$$(x+f(x))^{1/2} - x^{1/2} = 1/(2c_x^{1/2})\cdot f(x) \le f(x)/2x^{1/2}  \to 0.$$
Above $c_x \in (x,x+f(x)),$ so we're OK. In this problem $f(x) = (1+x^{1/2})^{1/2},$ so the desired limit is $0.$
