Limits of square root $$\lim_{x\to\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x + \sqrt x} }}-\sqrt x\right) $$
(original screenshot)
Compute the limit
Can you please help me out with this limit problem
 A: Hint: Try multiplying and dividing by the conjugate to get started, simplify the numerator, then factor $\sqrt x$ out of the new numerator and denominator.
A: Here's a comparatively clean way to do it: $$ \sqrt{x+\sqrt x}-\sqrt x\le\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}\le\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}-\sqrt{x} $$
Now, let $u=\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$.  Then $$u^2=x+u\implies u=\frac{1+\sqrt{1+4x}}{2}=\frac12+\sqrt{\frac14+x}$$
(Note that $u$ is strictly positive).  Now, $$\begin{align}\sqrt{x+\sqrt{x}}-\sqrt{x}&=\frac{x+\sqrt{x}-x}{\sqrt{x+\sqrt{x}}+\sqrt{x}}\\
&=\frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}\\
&=\frac{1}{\sqrt{1+\frac1{\sqrt x}}+1}\end{align}$$
Thus we have $$\begin{align} \lim_{x\to\infty}\frac{1}{\sqrt{1+\frac1{\sqrt x}}+1}\le\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}&\le\lim_{x\to\infty}\frac12+\sqrt{\frac14+x}-\sqrt x\\
\frac12\le\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}&\le\frac12\\
\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}&=\frac12 \end{align}$$
A: $$
\begin{align}
\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}-\sqrt{x}
&=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}}{\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}+\sqrt{x}}\\
&=\lim_{x\to\infty}\frac{\sqrt{1+\frac1x\sqrt{x+\sqrt{x+\dots}}}}{\sqrt{1+\frac1x\sqrt{x+\sqrt{x+\dots}}}+1}\\
&=\frac12
\end{align}
$$
To show that $\lim\limits_{x\to\infty}\frac1x\sqrt{x+\sqrt{x+\dots}}=0$, show inductively that
$$
\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}\le\frac{1+\sqrt{1+4x}}{2}
$$
using $\sqrt{x}\le\frac{1+\sqrt{1+4x}}{2}$ and
$$
\left(\frac{1+\sqrt{1+4x}}{2}\right)^2=x+\frac{1+\sqrt{1+4x}}{2}
$$
A: I got 1/2 for the limit. Let $y=\sqrt{x+\sqrt{x+\sqrt{x}}}$. $\frac{y}{\sqrt{x}} \rightarrow 1$ and $\frac{y}{x} \rightarrow 0$ as $x \rightarrow \infty$. 
And $L=\frac{\frac{y}{\sqrt{x}}}{\sqrt{1+\frac{y}{x}}+1} \rightarrow \frac{1}{2}$.
A: Hint for a simpler one:  $$\lim_{x \to \infty} \sqrt{x+\sqrt x}-\sqrt x=\lim_{x \to \infty}\sqrt x\left(\sqrt{1+\frac 1{\sqrt x}}-1\right)$$
A: Multiply the numerator and denominator by the conjugate expression. Divide the numerator and denominator by the greatest degree of $x$
