How do you evaluate this limit? $\lim\limits_{x\to-\infty}\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}$ $$\lim_{x\to-\infty}\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}$$
This is what I have tried so far, $$\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{10}+x^2)^{\frac{1}{2}}}$$
$$\begin{align}
&\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{10})^{\frac{1}{2}}+(x)^{\frac{1}{2}}}\\
&\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{5})+(x)}\\
&\lim_{x\to-\infty}\frac{x^2}{x}\\
&\lim_{x\to-\infty}x = -\infty\\
\end{align}$$
Is this the correct approach to finding the limit?

Well, if you can't get rid of a square root, put everything in a square root. Notice that for $x$ sufficiently negative--in particular, for $x<-\frac1{\sqrt[3]2}$--we have $2x^5+x^2<0,$ so that $$2x^5+x^2=-|2x^5+x^2|=-\sqrt{(2x^5+x^2)^2}$$ for sufficiently negative $x$-values. Thus, for such $x$-values, we have
$$\begin{align}
\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}
&= \frac{-\sqrt{(2x^5+x^2)^2}}{\sqrt{2x^{10}+x^2}} \\
&= -\sqrt{\frac{(2x^5+x^2)^2}{2x^{10}+x^2}} \\
&= -\sqrt{\frac{4x^{10}+4x^7+x^4}{2x^{10}+x^2}} \\
&= -\sqrt{2+\frac{4x^{10}+4x^7+x^4-2(2x^{10}+x^2)}{2x^{10}+x^2}} \\
&= -\sqrt{2+\frac{4x^7+x^4-2x^2}{2x^{10}+x^2}} \\
&= -\sqrt{2}\sqrt{1+\frac{4x^7+x^4-2x^2}{4x^{10}+2x^2}} \\
\end{align}
$$
The fraction inside the right-hand square root is about
$1/x^3$,
so the right-hand square root goes to $1$ as $x\to-\infty$,
so the whole thing goes to
$-\sqrt{2}$.
