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?
 A: For small $x$, $(1+x)^a \approx 1+ax$.  If you use this on both roots, the $1$s will cancel leaving you with a next term that is of order $\frac 1n$.  That takes care of the $n$ out front and you get a constant.
A: You were going on right track and shouldn't have stopped. There is also a typo when you take $2n$ out of square root. The terms in square root should be like $1+(1/4n)+\dots$.
Factoring out $2n$ allows us to rewrite the expression under limit as $$2n(\sqrt[3]{A}-\sqrt{B})$$ where $A, B$ are functions of $n$ tending to $1$ as $n\to\infty $. Now using this fact we add and subtract $1$ to get $$2n(\sqrt[3]{A}-1-(\sqrt{B}-1))$$ And this can be further rewritten as $$2n(A-1)\cdot\frac{\sqrt[3]{A}-1}{A-1}-2n(B-1)\cdot\frac {\sqrt{B} - 1}{B-1}$$ The fractions in above expression tend to $1/3,1/2$ respectively via the formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{1}$$ and observe that $$2n(A-1)=2n\left(\frac {1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}\right)\to 1$$ and similarly $$2n(B-1)\to \frac{1}{2}$$ The desired limit is thus $$1\cdot\frac{1}{3}-\frac{1}{2}\cdot\frac{1}{2}=\frac {1}{12}$$

Most usual limit problems related to algebraic functions don't need anything more than a little algebraic manipulation targeted  at making use of the limit formula $(1)$. Advanced tools like Taylor and L'Hospital's Rule are best left for difficult limit problems.
A: 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.)
A: Set $\dfrac1n=h,$
$$\lim_{n\to\infty}\sqrt[p]{2^pn^p+a_1n^{p-1}+\cdots+a_{p-1}n+a_p}-2n$$
$$=2\lim_{h\to0}\dfrac{\left(1+\dfrac{a_1}{2^p}h+\cdots+\dfrac{a_{p-1}}{2^p}h^{p-1}+\dfrac{a_p}{2^p}h^p\right)^{1/p}-1}h$$
$$=2\lim_{h\to0}\dfrac{1+\dfrac{a_1}{2^p}h+\cdots+\dfrac{a_{p-1}}{2^p}h^{p-1}+\dfrac{a_p}{2^p}h^p-1}h\cdot\dfrac1{\sum_{r=0}^{p-1}\left(1+\dfrac{a_1}{2^p}h+\cdots+\dfrac{a_{p-1}}{2^p}h^{p-1}+\dfrac{a_p}{2^p}h^p\right)^r}$$
$$=\dfrac{2a_1}{p2^p}$$
So, our required limit will be  $$\dfrac{2\cdot4}{3\cdot2^3}-\dfrac{2\cdot1}{2\cdot2^2}=\dfrac13-\dfrac14$$
A: $$
\begin{align*}
&\lim_{n\rightarrow \infty} \left( \sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9} \right) 
\\
&=\lim_{n\rightarrow \infty} 2n\left( \sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-\sqrt{1+\frac{1}{4n}+\frac{9}{4n^2}} \right) 
\\
&=\lim_{n\rightarrow \infty} 2n\left( \sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-1 \right) +\lim_{n\rightarrow \infty} 2n\left( 1-\sqrt{1+\frac{1}{4n}+\frac{9}{4n^2}} \right) 
\\
&=\lim_{n\rightarrow \infty} 2n\cdot \frac{1}{3}\left( \frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3} \right) -\lim_{n\rightarrow \infty} 2n\cdot \frac{1}{2}\left( \frac{1}{4n}+\frac{9}{4n^2} \right) 
\\
&=\frac{2}{3}\lim_{n\rightarrow \infty} \left( \frac{1}{2}+\frac{1}{8n}+\frac{11}{8n^2} \right) -\lim_{n\rightarrow \infty} \left( \frac{1}{4}+\frac{9}{4n} \right) 
\\
&=\frac{1}{3}-\frac{1}{4}
\\
&=\frac{1}{12}
\end{align*}
$$
