$\lim_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}$ 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.

Try looking at it this way: $$\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2}\cdot \sqrt{6-\frac{2}{x^2}}}{x+2}$$ Then since $x \rightarrow -\infty$ we have $x<0$ thus $\sqrt{x^2}=|x| = -x$ and we have 
$$\lim_{x\rightarrow -\infty} \frac{-x\cdot \sqrt{6-\frac{2}{x^2}}}{x+2}=\lim_{x\rightarrow -\infty} \frac{-\sqrt{6-\frac{2}{x^2}}}{1+\frac{2}{x}}=-\sqrt6$$ 
