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

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Compute the limit Can you please help me out with this limit problem

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closed as off-topic by Stefan Smith, Brian Rushton, Shuchang, Alex Wertheim, Dan Rust Dec 24 '13 at 0:25

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up vote 4 down vote accepted

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 $$\begin{align} \lim_{x\to\infty}\frac{1}{\sqrt{1+\frac1{\sqrt x}}+1}\le\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}&\le\lim_{x\to\infty}\frac12+\sqrt{\frac14+x}-\sqrt x\\ \frac12\le\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}&\le\frac12\\ \lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}&=\frac12 \end{align}$$

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Hint for a simpler one: $$\lim_{x \to \infty} \sqrt{x+\sqrt x}-\sqrt x=\lim_{x \to \infty}\sqrt x\left(\sqrt{1+\frac 1{\sqrt x}}-1\right)$$

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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.

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$$ \begin{align} \lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}-\sqrt{x} &=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}}{\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}+\sqrt{x}}\\ &=\lim_{x\to\infty}\frac{\sqrt{1+\frac1x\sqrt{x+\sqrt{x+\dots}}}}{\sqrt{1+\frac1x\sqrt{x+\sqrt{x+\dots}}}+1}\\ &=\frac12 \end{align} $$ 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 $$ \left(\frac{1+\sqrt{1+4x}}{2}\right)^2=x+\frac{1+\sqrt{1+4x}}{2} $$

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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 $L=\frac{\frac{y}{\sqrt{x}}}{\sqrt{1+\frac{y}{x}}+1} \rightarrow \frac{1}{2}$.

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Multiply the numerator and denominator by the conjugate expression. Divide the numerator and denominator by the greatest degree of $x$

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There is no numerator and no denominator. – MJD Dec 23 '13 at 21:53
@MJD: There is a denominator of $1$ that can be used for this purpose. – Ross Millikan Dec 24 '13 at 14:55

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