Help Evaluating $\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$ Does anyone know how to evaluate the following limit?
$$
\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}
$$
The answer is $\frac{1}{2}$, but I want to see a step by step solution if possible.
 A: $$\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x^{\frac{3}{2}}}}}+1}$$
I got the above by first multiplying by the conjugate to $\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}$ then dividing both the numerator and denominator by $\frac{1}{\sqrt{x}}$. Now take $x$ to infinity.
A: Using Taylor expansion:
$$\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$$
$$\sqrt{x + \sqrt{x}+\dfrac{1}{2} + O(\frac{1}{\sqrt{x}})} - \sqrt{x}$$
$$\sqrt{x}+\dfrac{1}{2\sqrt{x}}(\sqrt{x}+\dfrac{1}{2}+ O(\frac{1}{\sqrt{x}})) + O(\frac{1}{\sqrt{x}})- \sqrt{x}$$
$$\dfrac{1}{2} + \dfrac{1}{4\sqrt{x}}$$
Which in the limit case tends to ${\dfrac{1}{2}}$.
The expansion I used was $(a+b)^{1/2} = \sqrt{a}+\frac{b}{2\sqrt{a}} + \cdots$ 
A: $$
\begin{align}
\ \lim_{x\rightarrow\infty}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} &= \lim_{x\rightarrow\infty} \left( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} \right) \cdot \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\
\ &= \lim_{x\rightarrow\infty} \frac{x + \sqrt{x+\sqrt{x}}- x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\
\ &= \lim_{x\rightarrow\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} \\
\ &= \lim_{x\rightarrow\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} \cdot \frac{1/\sqrt{x}}{1/\sqrt{x}} \\
\ &= \lim_{x\rightarrow\infty} \frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\frac{1}{x}+\frac{1}{x\sqrt{x}}}+1} \\
\ &= \frac{\sqrt{1}}{\sqrt{1}+1} \\
\ &= \frac{1}{2} \\
\end{align}
$$
A: Multiply by the conjugate, then reduce:
\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) \cdot \frac{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\
&= \lim_{x\to\infty} \frac{(x + \sqrt{x + \sqrt{x}}) - x}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\
&= \lim_{x\to\infty} \frac{\sqrt{x + \sqrt{x}}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \cdot \frac{\dfrac{1}{\sqrt{x}}}{\dfrac{1}{\sqrt{x}}} \\
&= \frac{\sqrt{\lim\limits_{x\to\infty}\dfrac{1}{\sqrt{x}} + 1}}{\sqrt{1 + \lim\limits_{x\to\infty} \dfrac{\sqrt{x + \sqrt{x}}}{x}} + 1} \\
&= \frac{\sqrt{0 + 1}}{\sqrt{1 + 0} + 1} \\
&= \frac{1}{2}
\end{align*}
