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? 
 A: It is clear that our function $f(x)$ is negative when $x$ is negative.  Let $g(x)=(f(x)^2$. Is the value of
$$\lim_{x\to -\infty}g(x)$$
obvious? What can we conclude about our problem?
A: Noting that $x\lt 0$, we have$$\lim_{x\to -\infty}\frac{x}{\sqrt{x^2+2}}=\lim_{x\to -\infty}\frac{-(-x)}{\sqrt{x^2+2}}=\lim_{x\to -\infty}-\sqrt{\frac{(-x)^2}{x^2+2}}=\lim_{x\to -\infty}-\sqrt{\frac{1}{1+(2/x^2)}}=-1.$$
A: 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*}
A: Notice that in this case $\sqrt{x^{2}} = -x$, for $x<0$. Then 
$lim_{x\to - \infty}=\frac{x}{\sqrt{2+x^2}} = lim_{x \to -\infty} = \frac{x}{-x\sqrt{\frac{2}{x^2}+1}} = lim_{x \to -\infty}\frac{-1}{\sqrt{\frac{2}{x^2}+1}} = -1$
A: Here’s an answer that’s no better than those of André Nicolas and mathlove.
First, notice that $x=\text{sgn}(x)|x|$, where $\text{sgn}(a)$ is $-1$, $0$, or $1$ depending as $x$ is negative, zero, or positive. Second, notice that $|x|=\sqrt{x^2}$. So we have
$$
\frac x{\sqrt{x^2+2}}=\text{sgn}(x)\frac{\sqrt{x^2}}{\sqrt{x^2+2}}\,.
$$
Now combine the two radicals (which is all right to do, because both radicands are nonnegative), to get a rational expression inside, whose limit at $\pm\infty$ is $1$.
