<s>[INST] 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?
 [/INST] $\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 $v$.
</s>