Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-{\sqrt{4n^2+n+9}}$ Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9}$.
My first thoughts are the that I can remove a factor of $2n$ from each root, and obtain
$$ \lim_{n\to\infty}2n\sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-2n\sqrt{1+\frac{1}{2n}+\frac{9}{4n^2}} $$
But factoring out $2n$ won't help evaluate the limit because of the indeterminate form...  So then I thought of using L'Hop, but thinking about the chain rule there makes me think that its more effort than it's worth and there's a missing something I can't get to.  Any suggestions?

$$
\begin{align*}
&\lim_{n\rightarrow \infty} \left( \sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9} \right) 
\\
&=\lim_{n\rightarrow \infty} 2n\left( \sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-\sqrt{1+\frac{1}{4n}+\frac{9}{4n^2}} \right) 
\\
&=\lim_{n\rightarrow \infty} 2n\left( \sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-1 \right) +\lim_{n\rightarrow \infty} 2n\left( 1-\sqrt{1+\frac{1}{4n}+\frac{9}{4n^2}} \right) 
\\
&=\lim_{n\rightarrow \infty} 2n\cdot \frac{1}{3}\left( \frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3} \right) -\lim_{n\rightarrow \infty} 2n\cdot \frac{1}{2}\left( \frac{1}{4n}+\frac{9}{4n^2} \right) 
\\
&=\frac{2}{3}\lim_{n\rightarrow \infty} \left( \frac{1}{2}+\frac{1}{8n}+\frac{11}{8n^2} \right) -\lim_{n\rightarrow \infty} \left( \frac{1}{4}+\frac{9}{4n} \right) 
\\
&=\frac{1}{3}-\frac{1}{4}
\\
&=\frac{1}{12}
\end{align*}
$$
