Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^3$.
 A: Another way to solve the problem is the following, where no Differential Calculus is needed.
By elementary algebraic manipulation of radicals, you get:
$$ \begin{split} \sqrt[3]{n^3 + 3n^2} - \sqrt{n^2 + 2n} &= \underbrace{\sqrt[6]{(n^3 + 3n^2)^2}}_{=:\sqrt[6]{a}} - \underbrace{\sqrt[6]{(n^2 + 2n)^3}}_{=:\sqrt[6]{b}} \\ & = \frac{a - b}{\sqrt[6]{a^5} + \sqrt[6]{a^4b} + \sqrt[6]{a^3b^2} + \sqrt[6]{a^2b^3} + \sqrt[6]{ab^4} + \sqrt[6]{b^5}} \\ &\approx \frac{-3n^4}{6n^5} \\ &= - \frac{1}{2n} \end{split} $$
with the latter member going to $0$ as $n \to \infty$.
A: First let's get rid of the square root:
$$\lim_{n \to \infty} (\sqrt[3]{n^3-3n^2}-\sqrt{n^2+2n})=\lim_{n \to \infty} \frac{\sqrt[3]{(n^3+3n^2)^2}-(n^2+2n)}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}\\= -\lim_{n \to \infty} \frac{n^2+2n-\sqrt[3]{(n^3+3n^2)^2}}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}.$$
Now let's get rid of the cube root:
$$-\lim_{n \to \infty} \frac{n^2+2n-\sqrt[3]{(n^3+3n^2)^2}}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}$$
$$=-\lim_{n \to \infty} \frac{(n^2+2n)^3-(n^3+3n^2)^2}{(\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n})((n^2+2n)^2+(n^2+2n)\sqrt[3]{(n^3+3n^2)^2}+\sqrt[3]{(n^3+3n^2)^4})}.$$
The numerator yields $-3n^4$ and a bunch of lower-order terms, and the denominator yields $6n^5$ and a bunch of lower order terms, so dividing numerator and denominator by $n^5$ shows that the answer is $0$.
A: For this type of problem,
my preferred weapon
is the generalized binomial theorem
in the form
$(1+x)^a
=1+ax+a(a-1)x^2/2+O(x^3)
$.
Often the simpler form
$(1+x)^a
=1+ax+O(x^2)
$
is enough.
So,
$\begin{array}\\
\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n}
&=n\sqrt[3]{1+3/n}-n\sqrt{1+2/n}\\
&=n(1+(3/n)(1/3)+O(1/n^2))-n(1+(2/n)(1/2)+O(1/n^2))\\
&=n(1+1/n+O(1/n^2))-n(1+1/n+O(1/n^2))\\
&=n+1+O(1/n)-n-1-O(1/n)\\
&=O(1/n)\\
&\to 0
\qquad\text{as } n \to \infty\\
\end{array}
$
Note that
if what was wanted
was $n$ times the difference,
the additional term in the expansion
would have been needed.
You might try this
for practice.
A: Use the following fact: when $x\to 0$,
$$(1+x)^{\alpha}-1\sim\alpha x.$$
So 
$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n}=n\left(\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)\right)$$
$$=\frac{\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)}{\frac{1}{n}},$$
use the above fact:
$$\sqrt[3]{1+\frac{3}{n}}-1 \sim \frac{1}{n},\sqrt{1+\frac{2}{n}}-1\sim \frac{1}{n},$$
and this implies
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )=0.$$
In fact:
$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n}\sim -\frac{1}{2n},n\to\infty.$$
A: For positive $n$, let $a=\sqrt[3]{n^3+3n^2}$, and let $b=\sqrt{n^2+2n}$.

Note that $a < n+1$ and
\begin{align*}
3n +1 &= (n+1)^3-a^3\\[4pt]
&= \bigl((n+1)-a\bigr)\bigl((n+1)^2+a(n+1)+a^2)\\[4pt]
&> \bigl((n+1)-a\bigr)(n+1)^2\\[4pt]
\end{align*}
hence $0 < n+1-a < {\large{\frac{3n+1}{(n+1)^2}}}$, which implies $\displaystyle{\lim_{n\to \infty}}\;(n+1)-a = 0$.

Also, $b < n+1$ and
\begin{align*}
1&=(n+1)^2-b^2\\[4pt]
&= \bigl((n+1)-b\bigr)\bigl((n+1)+b\bigr)\\[4pt]
&> \bigl((n+1)-b\bigr)(n+1)\\[4pt]
\end{align*}
hence $0 < (n+1)-b < {\large{\frac{1}{n+1}}}$, which implies $\displaystyle{\lim_{n\to \infty}}\;(n+1)-b = 0$.

Then we get
$$\lim_{n\to \infty}\;a-b = \lim_{n\to \infty}\;\bigl(a - (n+1)\bigr) + \bigl((n+1)-b\bigr) = 0 + 0 = 0$$ 
A: There is a way to use directly the two binomial formulas you mentioned. It only needs a little trick:
$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} = \sqrt[3]{n^3+3n^2}\color{blue}{-n + n}-\sqrt{n^2+2n}$$
Now, consider
\begin{eqnarray*} \sqrt[3]{n^3+3n^2}-n
& \stackrel{n=\sqrt[3]{n^3}}{=} &  \frac{n^3+3n^2 - n^3}{(\sqrt[3]{n^3+3n^2})^2 + n\sqrt[3]{n^3+3n^2}+ n^2}\\& = & \frac{3}{\left(\sqrt[3]{1+\frac{3}{n}}\right)^2 + \sqrt[3]{1+\frac{3}{n}} + 1}\\
& \stackrel{n \to \infty}{\longrightarrow} & 1
\end{eqnarray*}
Similarly, you get $n - \sqrt{n^2+2n} \stackrel{n \to \infty}{\longrightarrow} -1$.
Hence, you get
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} ) = \lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-n ) + \lim_{n \to \infty} (n-\sqrt{n^2+2n} ) = 1-1 = 0$$
A: Set $1/x=h,h\to0^+$
The limit becomes $$\lim_{h\to0}\dfrac{\sqrt[3]{1+3h}-1}h-\lim...\dfrac{\sqrt{1+2h}-1}h$$
Now for $F=\lim_{...}\dfrac{\sqrt[m]{1+mh}-1}h,$
set $\sqrt[m]{1+mh}-1=u,$
$\implies mh=(1+u)^m-1=mu+O(u^2)$
Can you take it from here?
