<s>[INST] Find $\lim_{x\to-\infty}\frac{x}{\sqrt{x^2+2}}$ Find $\displaystyle \lim_{x\to-\infty}\frac{x}{\sqrt{x^2+2}}$ without a calculator.
L'Hopital's rule only gives me the limit for one side. Not two. How would I solve this problem? 
 [/INST] One definition of the absolute value function is $|x| = \sqrt{x^2}$.  For negative values of $x$, observe that
$$\frac{x}{|x|} = \frac{x}{-x} = -1$$
Thus,
\begin{align*}
\lim_{x \to -\infty} \frac{x}{\sqrt{x^2 + 2}} & = \lim_{x \to -\infty} \frac{x}{\sqrt{x^2\left(1 + \dfrac{2}{x^2}\right)}}\\
& = \lim_{x \to -\infty} \frac{x}{|x|\sqrt{\left(1 + \dfrac{2}{x^2}\right)}}\\
& = \lim_{x \to -\infty} \frac{x}{|x|} \lim_{x \to -\infty} \frac{1}{\sqrt{\left(1 + \dfrac{2}{x^2}\right)}}\\
& = -1 \cdot \frac{1}{\sqrt{1 + 0}}\\
& = -1 \cdot 1\\
& = -1
\end{align*}
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