How to solve $\lim _{x\to \infty }\left(\frac{\sqrt{3x+2}}{x\sqrt{x+6}-\sqrt{x^3+2x}}\right)$?

Question

I have a problem with this limit, i have no idea how to compute it. Can you explain the method and the steps used?

$$\lim _{x\to \infty }\left(\frac{\sqrt{3x+2}}{x\sqrt{x+6}-\sqrt{x^3+2x}}\right)$$

Answer 1

$$ \sqrt{3x+2}\sim \sqrt{3}\sqrt{x}+\frac{\sqrt{1/x}}{\sqrt{3}}-...$$ $$ x\sqrt{x+6}\sim x^{3/2}+3 \sqrt{x}-(9/2)\sqrt{1/x}+... $$ $$ \sqrt{x^3+2x}\sim x^{3/2}+\sqrt{1/x}+... $$ Therefore, the $x^{3/2}$ in the denominator cancels out, and what remains is $\sqrt{3x}/(3\sqrt{x})=1/\sqrt{3}$, which is the correct result.

Answer 2

Hint: $\frac{1}{\sqrt{a}-\sqrt{b}} = \frac{\sqrt{a}+\sqrt{b}}{a-b}$

Answer 3

$$\frac{\sqrt{3x+2}}{x\sqrt{x+6}-\sqrt{x^3+2x}}=\frac{\sqrt{3x+2}\left(x\sqrt{x+6}+\sqrt{x^3+2x}\right)}{x^2(x+6)-(x^3+2x)}$$

$$=\frac{\sqrt{3+\frac{2}{x}}\left(\sqrt{1+\frac{6}{x}}+\sqrt{1+\frac{2}{x^2}}\right)}{6-\frac{2}{x}}\stackrel{x\to \infty}\to \frac{\sqrt{3}\left(\sqrt{1}+\sqrt{1}\right)}{6}=\frac{1}{\sqrt{3}}$$

Answer 4

$$\begin{align}\lim _{x\to \infty }\left(\frac{\sqrt{3x+2}}{x\sqrt{x+6}-\sqrt{x^3+2x}}\right) &= \lim_{x\to\infty}\left[\left(\frac{\sqrt{3x+2}}{x\sqrt{x+6}-\sqrt{x^3+2x}}\right)\left({x\sqrt{x+6}+\sqrt{x^3+2x} \over x\sqrt{x+6}+\sqrt{x^3+2x} }\right)\right]\\ &= \lim_{x\to\infty}\left({\sqrt{3x+2}\left(x\sqrt{x+6}+\sqrt{x^3+2x}\right) \over x^2(x+6) - (x^3 + 2x) }\right) \\ &=\lim_{x\to\infty} \left({x\sqrt{(x+6)(3x+2)} + \sqrt{(x^3+2x)(3x+2)}\over6x^2-2x}\right) \\(*)& = \lim_{x\to\infty}\left({x\sqrt{3x^2} + \sqrt{3x^4}\over6x^2}\right) \\ &= \lim_{x\to\infty}\left({\sqrt3x^2 + \sqrt3x^2\over6x^2}\right)\\&= \lim_{x\to\infty}\left({2\sqrt3x^2\over6x^2}\right) \\& = \lim_{x\to\infty}\left({\sqrt3 \over 3}\right) \\ &= {\sqrt3 \over 3}\end{align}$$

The step $(*)$ is valid because at infinity, the smaller terms don't matter so we may discard them. The step immediately following is also valid because we are only working with positive values of $x$ as $x\to\infty$.