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$$\lim_{x \to \infty} \frac{\sqrt[3]{x+2}}{\sqrt{x+3}} $$

How would you proceed to find this limit, by eyeballing I would guess it foes to zero since the numerator has a smaller power than the denominator, normaly I would use the binomial theorem if I had something like $$\lim_{x \to \infty} \frac{\sqrt[3]{x+2}-1}{\sqrt{x+3}-1} $$ But here I don't know how to find the limit since I can't really use the binomial theorem.

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    $\begingroup$ For large $x$ the additive constants are neglectible and the expression tends to $x^{-1/6}$. $\endgroup$ – Yves Daoust Jan 13 '17 at 11:37
  • $\begingroup$ How would you use the binomial theorem? Because actually it is not a problem to add the -1 term, since it does not affect the limiting behavior of the fraction as $x\to \infty$. $\endgroup$ – Jimmy R. Jan 13 '17 at 11:47
  • $\begingroup$ I would put a = cubic root of (x+2) and b = 1 and choos n to be 3 so I would have (a^3-1^3)=(a-1)(a^2 + a^1 +1) so I can rewrite (a-1) as (a^3-1^3)/(a^2 + a^1 +1) and same for denominator $\endgroup$ – SoHCahToha Jan 13 '17 at 11:51
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If you factorize you get $$\frac{x^{1/3}(1+2/x)^{1/3}}{x^{1/2}(1+3/x)^{1/2}} = \frac{(1+2/x)^{1/3}}{x^{1/6}(1+3/x)^{1/2}} $$ I'll let you do the limit yourself.

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$$\lim_{x \to \infty} \frac{\sqrt[3]{x+2}}{\sqrt{x+3}}=\lim_{x \to \infty} \sqrt[6]{\dfrac{(x+2)^2}{(x+3)^3}}=\sqrt[6]{0}=0$$

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Simply use equivalents: $\;\sqrt[3]{x+2}\sim_\infty \sqrt[3]{x}$, $\;\sqrt{x+3}\sim_\infty \sqrt{x}$, hence $$\frac{\sqrt[3]{x+2}}{\sqrt{x+3}}\sim_\infty \frac{x^{1/3}}{x^{1/2}}=x^{-1/6}\xrightarrow[x\to\infty]{}0.$$

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  • $\begingroup$ I always look at your answers with awe as to what is equivalents ? Is it same as equivalence relations ? $\endgroup$ – A---B Jan 13 '17 at 20:50
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    $\begingroup$ It is an equivalence relation for functions defined near a point (possible near $\infty$). But in more detail, it means the ratio of the functions tends to $1$ (here at $\infty$). The main point is that it is compatible with multiplication and division and (with some restrictions) with composition by $\log$. It is a basic concept of Asymptotic analysis, together with notations $O$ and $o$. The ‘philosophy’ is to replace more or less complicated functions with simpler ones, so as to concentrate on the crux of the problem and delete irrelevant details. $\endgroup$ – Bernard Jan 13 '17 at 21:02
  • $\begingroup$ Do you work in this field of mathematics (It is analysis if I am correct) ? as I only seen you use this notation. $\endgroup$ – A---B Jan 13 '17 at 21:03
  • $\begingroup$ Sometimes other people use it, quite scarcely. Yes, I'm a mathematician, and, no, I'm an algebraist professionally. I learnt these notions during my 1st year after high school. $\endgroup$ – Bernard Jan 13 '17 at 21:07
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    $\begingroup$ You're welcome! $\endgroup$ – Bernard Jan 13 '17 at 21:40
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For $x \ge 2$ we have

$0 \le \frac{^3\sqrt{x+2}}{\sqrt{x + 3}} \le \frac{\sqrt[3]{2x}}{\sqrt{x}}$.

Your turn!

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A slightly longer way: use Generalized Binomial coefficients: $$ x^{-\frac{1}{6}}\frac{(1+\frac{2}{x})^{\frac{1}{3}}}{(1+\frac{3}{x})^{\frac{1}{2}}} \sim x^{-\frac{1}{6}}\frac{1+\frac{2}{x} + O(x^{-2})}{1+\frac{3}{x} + O(x^{-2}) } \to_x 0 $$

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Hint:

$$\frac{\sqrt[3]{x+2}}{\sqrt{x+3}}=\frac{\sqrt[3]x}{\sqrt x}\frac{\sqrt[3]{1+\frac2x}}{\sqrt{1+\frac3x}}.$$

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Hint: $x +2 < x + 3$.

(Yes, really, you can solve it using this.)

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  • $\begingroup$ @juniven hints are pretty allowed here. $\endgroup$ – djechlin Jan 14 '17 at 1:45

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