$$\lim_{x\to\infty} \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}}$$
I managed to find the limit of the above function, which according to my calculations is 1, but I don't know how to prove my answer is correct.
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asked Oct 12, 2013 at 14:00
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1$\begingroup$ Hint: Try dividing both the top and bottom by $\sqrt{x}$. Edit: I just realized you probably already did that. Do you want a $\delta-\epsilon$ proof? $\endgroup$– SuugakuOct 12, 2013 at 14:07
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2$\begingroup$ Examining the square of your expression may be easier. $\endgroup$– David MitraOct 12, 2013 at 14:14
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2$\begingroup$ Come on Diane, you're kidding. Rewrite your expression to $$\sqrt{\frac{x+ \sqrt{ x+\sqrt x}}{x+1}}$$ and then rduce by $x$. $\endgroup$– Michael HoppeOct 12, 2013 at 14:15
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1$\begingroup$ @MichaelHoppe I don't know how to prove my answer is correct meaning I don't know how to write a formal proof to show that 1 is the limit of that function. $\endgroup$– Diane VanderwaifOct 12, 2013 at 14:18
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1$\begingroup$ @DianeVanderwaif We don't know either because we don't know what you're allowed to use. For example if you know about continuous functions just follow my comment and you're done. In case you have to use the $\epsilon$-$\delta$-thing I wish you good luck. I'm too old for those nasty things. $\endgroup$– Michael HoppeOct 12, 2013 at 14:29
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6 Answers
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Since $$\sqrt{x}=_{\infty}o(x)$$ then simply we have $$ \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}}\sim_\infty \frac{\sqrt x}{\sqrt x}=1$$
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2$\begingroup$ I dont know why the asymptotic notations during the evaluations of limits doesnt come out more often. It's so easy and elegant! $\endgroup$– AntOct 12, 2013 at 18:12
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If you want a definition based proof ($M-\epsilon$) just note that for large $x$: $$ \left|\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}-1\right|=\\ \left|\frac{x+\sqrt{x}-1}{\sqrt{x+\sqrt{x}}\cdot \left(\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x+1}\right)\cdot\left(\sqrt{x+\sqrt{x}}+1\right)}\right|\leq\frac{2x}{x\sqrt{x}}=\frac{2}{\sqrt{x}}. $$
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For $x>0$ we have $$ \frac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt{x+\sqrt x}}=\frac{\sqrt x\cdot \sqrt{1+\frac{\sqrt{x+\sqrt x}}x}}{\sqrt x\cdot\sqrt{1+\frac1{\sqrt x}}} =\frac{\sqrt{1+\frac1{\sqrt x}\sqrt{1+\frac1{\sqrt x}}}}{\sqrt{1+\frac1{\sqrt x}}}.$$ Now as $x\to\infty$, all those $\frac1{\sqrt x}$ tend to $0$, hence all $\sqrt {1+\ldots}$ tend to $1$, hence so does the fraction.
answered Oct 12, 2013 at 14:20
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If you replace X by 1 / Y^2 and make a Taylor expansion around Y = 0, you should get it. I hope this is of some help to you.
answered Oct 12, 2013 at 14:13
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Yes. Your answer is correct because $$ \lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x(1+\frac{\sqrt{x}}{x}}})}{\sqrt{x(1+\frac{1}{x})}}; $$
Because $\lim_{x\to\infty}\frac{1}{x}=0$, and $\lim_{x\to\infty}\frac{\sqrt{x}}{x}=0$ we have:
$$ \lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x(1+\frac{\sqrt{x}}{x}}})}{\sqrt{x(1+\frac{1}{x})}}=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x}}=\lim_{x\to\infty}\frac{\sqrt{x(1+\frac{\sqrt{x}}{x}})}{\sqrt{x}}=\lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x}}=1 $$
answered Oct 12, 2013 at 14:14
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$\begingroup$ that' how I found my answer, what I need is a proof to show that 1 is limit of that $\endgroup$ Oct 12, 2013 at 14:22
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1$\begingroup$ @Madrit Zhaku: how would you rigorously justify the second and fourth equalities in your chain? Replacing a sub-expression under a complex limit by the limit of the sub-expression is not in general kosher. (That is: it’s not always the case that $\lim_{x \to \infty} f(x,g(x)) = \lim_{x \to \infty} f(x,\lim_{x \to \infty} g(x))$.) I’m not saying that those steps are wrong, but I think they need a bit more justification. $\endgroup$ Oct 12, 2013 at 17:27
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$\begingroup$ @PeterLeFanuLumsdaine, a justification can be --- when $f$ is continuous, it's true that $\lim f( g(x) ) = f( \lim g(x))$. In the context of this problem, it seems reasonable to use the fact that $\sqrt{x}$ is continuous. It's also easy to prove that $\sqrt{x}$ is continuous and then use the fact. I don't know how easy it is to prove that the continuity of $f$ implies the result mentioned, but I do believe someone like James Stewart proves that at least in an an appendix. The OP did not specify how he wanted a proof. $\endgroup$ Sep 28 at 17:53
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That's obious: $$ \frac {\sqrt{x+ \sqrt{ x+\sqrt x}} }{\sqrt{x+1}} =\sqrt{\frac{x+ \sqrt{ x+\sqrt {x}}}{x+1}} =\sqrt{\frac{1+\sqrt{1/x+1/\sqrt{x}}}{1+1/x}}.$$
answered Oct 12, 2013 at 18:09