Evalute $\lim_{x\to-\infty} \frac{\sqrt{x^2+4x^4}}{8+x^2}$ Having a hard time with this. So far I have:
$$ \frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \frac{x\sqrt{1+4x^2}}{8+x^2}$$

I am assuming that you mean
$\frac{\sqrt{x^2+4x^4}}{8+x^2}$.
Since $x^2$ and $x^4$ are positive,
the limit is the same for
$x \to \infty$ and
$x \to -\infty$.
You can take your last step
further and write
$\begin{align}
\frac{\sqrt{x^2+4x^4}}{8+x^2}
&= \frac{x\sqrt{1+4x^2}}{8+x^2}\\
&= \frac{x^2\sqrt{1/x^2+4}}{x^2(1+8/x^2)}\\
&= \frac{\sqrt{1/x^2+4}}{1+8/x^2}\\
\end{align}
$
As $x \to \pm \infty$,
$\sqrt{1/x^2+4} \to \sqrt{4} = 2$
and
$1+8/x^2 \to 1$,
so I get
$\frac{2}{1} = 1$.
