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I saw some resolutions here like $\sqrt{x+\sqrt{x+\sqrt{x}}}- \sqrt{x}$, but I couldn't get the point to find $\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.

I tried $\frac{1}{x}.(\sqrt{x+\sqrt{x+\sqrt{x}}})=\frac{\sqrt{x}}{x}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)=\frac{1}{\sqrt{x}}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)$ but now I got stuck. Could anyone help?

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4 Answers 4

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For large $x$ you have $\sqrt x<x$ and so $$ \sqrt{x+\sqrt{x+\sqrt{x}}}<\sqrt{x+\sqrt{2x}}<\sqrt{3x}. $$ Since $\sqrt{x+\sqrt{x+\sqrt{x}}}/x<\sqrt{3x}/x\to0$ when $x\to\infty$, your limit is zero.

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$\dfrac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x} = \sqrt{\dfrac{x+\sqrt{x+\sqrt{x}}}{x^2}}\\ = \sqrt{\dfrac{1}{x} + \dfrac{\sqrt{x+\sqrt{x}}}{x^2}} = \sqrt{\dfrac{1}{x} + \sqrt{\dfrac{1}{x^3}+\sqrt{\dfrac{1}{x^7}}}}$

then the limit when $x \rightarrow \infty$ is clearly $0$

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  • $\begingroup$ Much simpler :) $\endgroup$
    – Spencer
    Jan 28, 2016 at 3:16
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Jonas produced the solution first, but I wanted to write up a solution for my own edification.


For $x>1$ we can write the following inequalities,

$$x < x+a\sqrt{x} < (1+a)x\qquad (a>0),$$

This allows us to come up with an upper and lower bound for the numerator,

$$\sqrt{x+\sqrt{x+\sqrt{x}}} < \sqrt{x+\sqrt{2x}}< \sqrt{(1+\sqrt{2})x} = \sqrt{x}\ \sqrt{(1+\sqrt{2})}$$

$$\sqrt{x+\sqrt{x+\sqrt{x}}} > \sqrt{x+\sqrt{x}} > \sqrt{x}$$

So our function, which we are taking the limit of, has the following bounds on it,

$$\frac{\sqrt{x}}{x} <\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}<\frac{\sqrt{x}}{x}\sqrt{1+\sqrt{2}}, $$

as $x\rightarrow \infty$ the function $\sqrt{x}/x$ goes to $0$. Our function is bounded above by something which goes to zero and below by something which goes to zero it must also go to zero; this is called the squeeze theorem.

$$\lim_{x\rightarrow \infty} \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x} = 0.$$

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  • $\begingroup$ If anyone can tell me a good name with which to refer to $f(x)$ in the expression $\lim_{x \rightarrow a} f(x)$ I would appreciate it. Integrand is such a useful term, there should be something similar for other operations. $\endgroup$
    – Spencer
    Jan 28, 2016 at 0:16
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For an alternative answer, consider the infinitely nested radical expression $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}}$.

Evidently $y > \sqrt{x+\sqrt{x+\sqrt{x}}}$ for $x > 1$ and therefore $\frac{y}{x}$ is an upper bound for $\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.

Also $y > \sqrt{x}$ and therefore $\lim_{x\to\infty} y = \infty$

But we can express $x$ in terms of $y$ by using the substitution $y = \sqrt{x+y}$ in the infinitely nested expression, leading to $x = y(y-1)$.

Then $\lim_{x\to\infty}\frac{y}{x} = \lim_{y\to\infty} \frac{y}{y(y-1)} = \lim_{y\to\infty}\frac{1}{y-1} = 0$.

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