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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.

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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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  • $\begingroup$ I like your answer! :) $\endgroup$ – dmk Jun 15 '14 at 1:46
  • $\begingroup$ thanks i like ur answer to $\endgroup$ – user157128 Jun 15 '14 at 1:54
  • $\begingroup$ as you did in step 3 to 4 when multiplied by 1/v5 ??? $\endgroup$ – user157128 Jun 15 '14 at 2:03
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$$\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x^{\frac{3}{2}}}}}+1}$$

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.

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  • $\begingroup$ the answer is 1/2, but I tried this and rationalize, but did not reach 0.5 $\endgroup$ – user157128 Jun 15 '14 at 1:38
  • $\begingroup$ @user157128 what did you get? In the top I get $\sqrt{1 + 0} - 0$. What do you get in the bottom? $\endgroup$ – DanZimm Jun 15 '14 at 1:40
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$$ \begin{align} \ \lim_{x\rightarrow\infty}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} &= \lim_{x\rightarrow\infty} \left( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} \right) \cdot \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\ \ &= \lim_{x\rightarrow\infty} \frac{x + \sqrt{x+\sqrt{x}}- x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\ \ &= \lim_{x\rightarrow\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} \\ \ &= \lim_{x\rightarrow\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} \cdot \frac{1/\sqrt{x}}{1/\sqrt{x}} \\ \ &= \lim_{x\rightarrow\infty} \frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\frac{1}{x}+\frac{1}{x\sqrt{x}}}+1} \\ \ &= \frac{\sqrt{1}}{\sqrt{1}+1} \\ \ &= \frac{1}{2} \\ \end{align} $$

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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 $(a+b)^{1/2} = \sqrt{a}+\frac{b}{2\sqrt{a}} + \cdots$

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  • $\begingroup$ It's not clear from what you've written why you get to ignore the error terms in your expansion. $\endgroup$ – Cheerful Parsnip Jun 15 '14 at 1:50
  • $\begingroup$ @user157107 this seems like a fairly interesting way to go about it... and similar to what I was considering doing. Could you possibly add more details to justify your approach? $\endgroup$ – Squirtle Jun 15 '14 at 2:13

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