<s>[INST] 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.
 [/INST] 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*}
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