# Limit at infinity of cubic roots and square roots without using conjugate $\lim_{x \to \infty} \frac{\sqrt[3]{x+2}}{\sqrt{x+3}}$

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

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.

• For large $x$ the additive constants are neglectible and the expression tends to $x^{-1/6}$.
– user65203
Jan 13 '17 at 11:37
• 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 $x\to \infty$. Jan 13 '17 at 11:47
• 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 Jan 13 '17 at 11:51

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.

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

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

• I always look at your answers with awe as to what is equivalents ? Is it same as equivalence relations ?
– user312097
Jan 13 '17 at 20:50
• 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. Jan 13 '17 at 21:02
• Do you work in this field of mathematics (It is analysis if I am correct) ? as I only seen you use this notation.
– user312097
Jan 13 '17 at 21:03
• 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. Jan 13 '17 at 21:07
• You're welcome! Jan 13 '17 at 21:40

For $x \ge 2$ we have

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

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

Hint:

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

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

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

• @juniven hints are pretty allowed here. Jan 14 '17 at 1:45