Limit as $x\to -\infty$ for function having square root

Question

Evaluate $$\lim_{x\to -\infty} \frac{\sqrt{3x^2+2}}{x-2}.$$

My work: when I solved $$\lim_{x\to -\infty} \frac{\sqrt{3+2/x^2}}{1-2/x}$$ I got answer $\sqrt{3}$ but if we take $|x|$ then I got $-\sqrt{3}$ what should be answer ? $\sqrt{3}$ or $-\sqrt{3}$ ?

Answer 1

$y:=-x.$

Now consider:

$\lim_{y \rightarrow \infty}\dfrac{\sqrt{3y^2 +2}}{-y-2}$

Answer 2

Note that $\sqrt{x^2}=|x|$ and, for $x<0$, $|x|=-x$. Therefore, as $x\to -\infty$, $$\frac{\sqrt{3x^2+2}}{x-2}=\frac{\sqrt{x^2}\sqrt{3+\frac{2}{x^2}}}{x(1-\frac{2}{x})}=\frac{|x|\sqrt{3+\frac{2}{x^2}}}{x(1-\frac{2}{x})}=-\frac{\sqrt{3+\frac{2}{x^2}}}{1-\frac{2}{x}}\to -\sqrt{3}.$$

Answer 3

write $$\lim_{x\to -\infty}\frac{|x|}{x}\frac{\sqrt{3+\frac{2}{x^2}}}{1-\frac{2}{x}}$$