How to find limit of $\lim_{n\to\infty} (\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$)? I am stuck on this limit. 
$$\lim_{n\to\infty} \sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$$
I couldn't find the limit using the basic properties of limits, since that just yields: $$\infty-\infty$$ which is undefined. Could I get any hints for finding this limit?
 A: Hint: Mulitply by
$$
\frac{\left(\sqrt[3]{n^2+5}\right)^2 + \sqrt[3]{n^2+5}\sqrt[3]{n^2+3} + \left(\sqrt[3]{n^2+3}\right)^2}{\left(\sqrt[3]{n^2+5}\right)^2 + \sqrt[3]{n^2+5}\sqrt[3]{n^2+3} + \left(\sqrt[3]{n^2+3}\right)^2}
$$
A: You can write $\sqrt[3]{1+x}=1+\frac{1}{3}x+O(x^2)$ for $x$ near zero.
So $$\sqrt[3]{n^2+3}=\sqrt[3]{n^2}\cdot \sqrt[3]{1+3/n^2} = n^{2/3}+\frac{1}{3}\frac{3}{n^{4/3}} + O(1/n^{10/3})$$
Similarly, $$\sqrt[3]{n^2+5}=\sqrt[3]{n^2}\cdot \sqrt[3]{1+5/n^2} = n^{2/3}+\frac{1}{3}\frac{5}{n^{4/3}} + O(1/n^{10/3})$$
So $$\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}=\frac{2}{3}n^{-4/3}+O(n^{-10/3})$$

Alternatively, you can use the mean value theorem. Let $f(x)=\sqrt[3]{x}$. Then for any $n$ there is a $c_n\in [n^2+3,n^2+5]$ such that:
$$f(n^2+5)-f(n^2+3)=((n^2+5)-(n^2+3))f'(c_n)=2f'(c_n)$$
Now, show that $0< f'(c_n)\leq f'(n^2+2)$ and $f'(n^2+2)\to 0.$
A: One developes an intuition that as $n^2$ becomes large the $+5$ and $-3$ become negligible and the whole thing becomes but how to prove it...
Well we can is eliminate the difference of square roots by multiplying by the compliment so we can probably do the same for cube roots.
I.e. $(\sqrt[3]a - \sqrt[3]b)(\sqrt[3]a^2 + \sqrt[3]a\sqrt[3]b+\sqrt[3]b^2) = a - b$.
So $\lim (\sqrt[3]{n^2 + 5} - \sqrt[3]{n^2 + 3})=$
$\lim (\sqrt[3]{n^2 + 5} - \sqrt[3]{n^2 + 3})\frac {\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}{\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}=$
$\lim \frac{(n^2 + 5)-(n^2 +3)}{\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}=$
$\lim \frac{2}{\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}=0$
A: $\lim_{n\to\infty} \sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}
$
Since
$a^3-b^3
=(a-b)(a^2+ab+b^2)
$,
$a-b
=\dfrac{a^3-b^3}{a^2+ab+b^2}
$.
Therefore
$a^{1/3}-b^{1/3}
=\dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}
$.
Since,
for $x > 0$,
$1 < (1+x)^{1/3}
< 1+x/3
$
(cube both sides),
we have,
if $u > 0$,
$n^{2/3}
< (n^2+u)^{1/3}
= n^{2/3}(1+u/n^2)^{1/3}
< n^{2/3}(1+u/(3n^2))
= n^{2/3}+u/(3n^{4/3}))
$.
Therefore,
if $a = n^2+u$
and $b = n^2+v$,
$\begin{array}\\
3n^{4/3}
&< a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\\
&< (n^{2/3}+u/(3n^{4/3})))^2+(n^{2/3}+u/(3n^{4/3})))(n^{2/3}+v/(3n^{4/3})))+(n^{2/3}+v/(3n^{4/3})))^2\\
&= n^{4/3}( (1+u/(3n^2))^2+(1+u/(3n^2))(1+v/(3n^2))+(1+v/(3n^2))^2)\\
&= n^{4/3}(3+(2u+u+v+2v)/(3n^2)+(u^2+uv+v^2)/(3n^2)^2\\
&= 3n^{4/3}(1+(u+v)/(3n^2)+(u^2+uv+v^2)/(3n^4))\\
\end{array}
$
Therefore
$$a^{1/3}-b^{1/3}
=\dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}
< \dfrac{u-v}{3n^{4/3}}
$$
and
$$a^{1/3}-b^{1/3}
> \dfrac{u-v}{3n^{4/3}(1+(u+v)/(3n^2)+(u^2+uv+v^2)/(3n^4))}.
$$
Immediately we have that
$\lim_{n \to \infty} a^{1/3}-b^{1/3}
=0
$.
More precisely,
we also have
$$\dfrac{u-v}{3}
> n^{4/3}(a^{1/3}-b^{1/3})
>\dfrac{u-v}{3}\dfrac1{1+(u+v)/(3n^2)+(u^2+uv+v^2)/(3n^4)}
$$
so that
$$\lim_{n \to \infty} n^{4/3}(a^{1/3}-b^{1/3})
=(u-v)/3
=2/3
$$
since
$u=5, v=3$.
Note:
We can get explicit bounds
by using
$\dfrac1{1+x}
\gt 1-x
$
for $0 < x < 1$
and choosing
$n$ large enough
compared with
$u$ and $v$.
A: If you have the Mean Value Theorem available, and if you know that the derivative of $f(x)=x^{1/3}$ is $f'(x)={1\over3}x^{-2/3}$, then note that
$$\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}=2{f(n^2+5)-f(n^2+3)\over(n^2+5)-(n^2+3)}=2f'(c_n)={2\over3c_n^{2/3}}$$
for some $c_n\in(n^2+3,n^5+5)$.  Clearly $c_n\to\infty$ as $n\to\infty$, so we get
$$\lim_{n\to\infty}(\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3})=\lim_{n\to\infty}{2\over3c_n^{2/3}}=0$$
