<s>[INST] Limit as $x\to -\infty$ for function having square root 
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 $-\sqrt{3}$ ?
 [/INST] Note that $\sqrt{x^2}=|x|$ and, for $x<0$, $|x|=-x$. Therefore, as $x\to -\infty$,
$$\frac{\sqrt{3x^2+2}}{x-2}=\frac{\sqrt{x^2}\sqrt{3+\frac{2}{x^2}}}{x(1-\frac{2}{x})}=\frac{|x|\sqrt{3+\frac{2}{x^2}}}{x(1-\frac{2}{x})}=-\frac{\sqrt{3+\frac{2}{x^2}}}{1-\frac{2}{x}}\to -\sqrt{3}.$$
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