Limit of √(x+√(x+√x))-√x as x approaches infinity

[anemone]

Evaluate the limit, if it exists:

$\displaystyle \lim_{{x}\to{\infty}} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$.

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[anemone]

Congratulations to the following members for their correct solutions:

1. Ackbach
2. magneto

Here is Ackbach's solution;

We have:

Spoiler

\begin{align*}
\lim_{x\to\infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} \right)
&=\lim_{x\to\infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} \right) \left(\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\right) \\
&=\lim_{x\to\infty}\left(\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\right) \\
&=\lim_{x\to\infty}\left(\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\right) \\
&=\lim_{x\to\infty}\left(\frac{\sqrt{1+\sqrt{\frac1x}}}{\sqrt{1+\sqrt{\frac1x+\sqrt{\frac{1}{x^3}}}}+\sqrt{1}}\right) \\
&=\frac12.
\end{align*}

Here is the solution submitted by magneto, who approached the problem a bit different than Ackbach:

Spoiler

Let $x = \frac 1{h^2}$, and take the limit as $h \to 0$. The expression then simplify:

\[
\sqrt{\frac 1{h^2} + \sqrt{\frac 1{h^2} + \frac 1h}} - \frac 1h = \sqrt{\frac 1{h^2} + \frac 1h \sqrt{1+h}} - \frac 1h =
\frac 1h \left( \sqrt{1 + h \sqrt{1+h}} - 1 \right)
\]

Multiply by the conjugate, we have:
\[
\frac{1 + h\sqrt{1+h} - 1}{h \left( \sqrt{1 + h \sqrt{1+h}} + 1 \right)} = \frac{\sqrt{1+h}}{ \left( \sqrt{1 + h \sqrt{1+h}} + 1 \right)} \to \frac 12,
\text{ as $h \to 0$.}
\]