<s>[INST] Solution verification: $\lim\limits_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} = -1$ I am trying to find the following limit 
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2}$$
and I did the following steps: 
\begin{align}
\require{cancel}
&\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} \\
&\lim_{x\rightarrow \infty} \frac{x^2\left(\frac{\sqrt{x}}{x^2}+1\right)}{x^2\left(\frac{2x}{x^2}-1\right)} \\
& \lim_{x\rightarrow \infty} \frac{\cancel{x^2}\left(\frac{\sqrt{x}}{x^2}+1\right)}{\cancel{x^2}\left(\frac{2\cancel{x}}{\cancel{x^2}}-1\right)}\\
& \lim_{x\rightarrow \infty} \frac{\left(\frac{\sqrt{x}}{x^2}+1\right)}{\left(\frac{2}{x}-1\right)} \\
\end{align}
Now here, the top portion goes to $0$ because the there is a larger power of $x$ in the denominator leaving only a $+1$ on top. On the bottom, the same thing happens, $\frac{2}{x}$ goes to $0$ and we left with $-1$ in the denominator. Therefore
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} = -1$$
Is my solution correct and did I take the right steps with the correct logic?
Thanks!
 [/INST] Looks good! I would be a bit more precise about "the top portion goes to $0$," but your reasoning is just fine.
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