<s>[INST] $\lim_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}$ May I know how can I calculate the following expression?
$$
\lim\limits_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}
$$
From calculator, the answer is $-\sqrt{6}$ , my approach is by using dividing numerator and denominator by using $x$. Which is, 
$$
\lim\limits_{x\to-\infty}\frac{\frac{\sqrt{6x^2 - 2}}{x}}{\frac{x+2}{x}}
=
\lim\limits_{x\to-\infty}\frac{{\sqrt{\frac{6x^2 - 2}{x^2}}}}{\frac{x+2}{x}}=\sqrt{6}
$$
My answer is $\sqrt{6}$, is my working wrong or there are actually another approach? Thank you.
 [/INST] $$\lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{x+2}$$
First note that as $x\to -\infty$, then $x+2\lt 0$ and
$$ x+2=-|x+2|=-\sqrt{(x+2)^2} $$
Therefore 
$$ \lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{x+2} = \lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{-\sqrt{(x+2)^2}} $$
$$ = -\lim_{x\to -\infty} \sqrt{\frac{6x^2-2}{x^2+4x+4}} = -\lim_{x\to -\infty} \sqrt{\frac{6-\frac{2}{x^2}}{1+\frac{4}{x}+\frac{4}{x^2}}} $$
$$ =-\sqrt{\frac{6-0}{1+0+0}}=-\sqrt{6} $$
</s>