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I was given the following problem : $$\lim \limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}$$

The following is my approach:

$= \sqrt{ \lim \limits_{x \to \infty} \frac{x}{x+\sqrt{x+\sqrt{x+...}}} } $

I divide by x:

$= \sqrt{ \lim \limits_{x \to \infty} \frac{1}{1+\frac{\sqrt{x+\sqrt{x+...}}}{x}} } $

I then reasoned that obviously: $ \sqrt{x+\sqrt{x+...}}<x$
that is $ \frac{\sqrt{x+\sqrt{x+...}}}{x} = \frac{1}{x^y} : y>0 $
hence $\lim \limits_{x \to \infty} \frac{\sqrt{x+\sqrt{x+...}}}{x} =0$

I concluded that the original limit equals $ \sqrt{\frac{1}{1+0}} = 1$

Was my reasoning correct? And is this a legitimate mathematical approach to solve this limit?

Ahmed Elyamani
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  • You should use \cdots for dots between plus signs, use \dots for dots between commas. – Gregory Grant May 30 '15 at 15:45
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    The denominator has a closed form if you are interested: http://mathworld.wolfram.com/NestedRadical.html – muaddib May 30 '15 at 15:47
  • Going from $\sqrt{x + \sqrt{x + \cdots}} < x$ to deciding that it must be polynomial in $x$ is invalid – ruler501 May 30 '15 at 15:48
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    Actually, there's a simpler approach for this: Let $y=\sqrt{x+\sqrt{x+...}}$, $y^2-x=y$, $\sqrt x=\sqrt{y^2-y}$. So $\lim \limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}=\lim\limits_{y\to\infty}\frac{\sqrt {y^2-y}}{y}=\lim\limits_{y\to\infty}\sqrt{1-\frac{1}y}=1$ – Mythomorphic May 30 '15 at 15:51
  • The "obviously" part needs to be proved, in fact you assume without proof more, that the limit of the ratio is $0$. – André Nicolas May 30 '15 at 15:55

1 Answers1

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You could have taken this approach as well:

$$A=\sqrt{x+\sqrt { x+\sqrt {x+...}}}$$

$$A^2=x+\sqrt{x+\sqrt { x+\sqrt {x+...}}}=x+A$$

$$A^2-A=x$$

$$A^2-A+\frac14=x+\frac14$$

$$(A-\frac12)^2=x+\frac14$$

$$A=\sqrt{x+\frac14}+\frac12$$

then the limit becomes

$$\lim \limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x+\frac14}+\frac12}$$

Which is indeed 1

Gappy Hilmore
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