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

If $x\lt 0$, then
$$\frac{\sqrt{6x^2-2}}{x}\not=\sqrt{\frac{6x^2-2}{x^2}}$$
because the left-hand-side is negative and the right-hand-side is positive.
Since $-x\gt 0\Rightarrow -x=\sqrt{(-x)^2}=\sqrt{x^2}$, we have
$$\frac{\sqrt{6x^2-2}}{x}=\frac{\sqrt{6x^2-2}}{-(-x)}=\frac{\sqrt{6x^2-2}}{-\sqrt{(-x)^2}}=-\sqrt{\frac{6x^2-2}{(-x)^2}}=-\sqrt{\frac{6x^2-2}{x^2}}.$$
