<s>[INST] 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?
 [/INST] The expression seems designed for sheer cussedness if one attempts many of the common methods, since the indices of the roots differ so the "conjugate-factor" method is not much of an option.  (And one should avoid applying LHR to expressions with sums or differences of radicals unless one enjoys suffering...)
An approximation of some sort is more helpful. Ross Millikan's approach makes use of the "binomial approximation".  Another way that is accessible, since roots of suitable polynomials are being taken, is to "complete" the cubes and squares:
$$ \lim_{n \ \rightarrow \ +\infty} \ \sqrt[3]{8n^3+4n^2+n+11} \ - \ {\sqrt{4n^2+n+9}}    $$
$$ = \ \lim_{n \ \rightarrow \ +\infty} \ \ 2 \ \sqrt[3]{\left( n^3 \ + \ \frac12 \ n^2 \ + \ \frac18 \ n \ + \ \frac{11}{8} \right)} \ - \ 2 \ {\sqrt{\left(n^2 \ + \ \frac14 \ n \ + \ \frac94 \right)}} $$
$$ = \ \lim_{n \ \rightarrow \ +\infty} \ \ 2 \ \sqrt[3]{  \ \left[ n^3 \ + \ 3·\frac16 \ n^2 \ + \ 3·\frac{1}{36} \ n \ + \ \frac{1}{216}
 \right]+ \ \left(\frac18 - \frac{1}{12} \right)  n \ + \ \left(\frac{11}{8} - \frac{1}{216} \right)} \ - \ 2 \ \sqrt{\left[n^2 \ + \ 2·\frac18 \ n \ + \ \frac{1}{64} \right] \ + \ \left( \frac94 - \frac{1}{64}  \right)} $$
$$ = \ \lim_{n \ \rightarrow \ +\infty} \ \ 2 \ \sqrt[3]{  \ \left( n \ + \ \frac16 \right)^3 \ + \  \left[ \ \text{small terms} \ \right]} \ \ - \ \ 2 \ \sqrt{ \ \left(n  \ + \  \frac18  \right)^2 \ + \  \left[ \ \text{small terms} \ \right]} $$
$$ \approx \ \lim_{n \ \rightarrow \ +\infty} \ \ 2· \left( n \ + \ \frac16 \right) \ \ - \ \ 2 · \left| \ n  \ + \  \frac18 \ \right|  \ \ , $$
which agrees with the "binomial approximation" result.  (One has to magnify a graph a fair bit to see that the limit is not zero.)
Incidentally, this also shows that the horizontal asymptote for this function is "one-sided".  We observe that the asymptote corresponding to the limit "at negative infinity" for this expression is a line of slope $ \ 4 \ \ , $ giving us $$ \lim_{n \ \rightarrow \ -\infty} \ \sqrt[3]{8n^3+4n^2+n+11} \ - \ {\sqrt{4n^2+n+9}}  \ \ = \ \ -\infty \ \ .  $$
(We can consider this situation since $ \ 4x^2+x+9 > 0 \ $ for all real numbers.)
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