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

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

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 $$\lim_{x \to \infty} \frac{\sqrt{x+2}-1}{\sqrt{x+3}-1}$$ 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 $x^{-1/6}$. – Yves Daoust 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 $x\to \infty$. – Jimmy R. 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 – SoHCahToha Jan 13 '17 at 11:51

## 7 Answers

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.

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

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

• I always look at your answers with awe as to what is equivalents ? Is it same as equivalence relations ? – A---B 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. – Bernard 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. – A---B 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. – Bernard Jan 13 '17 at 21:07
• You're welcome! – Bernard Jan 13 '17 at 21:40

For $x \ge 2$ we have

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

Your turn!

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:

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

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

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

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