Question: 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?

Answer: 
Hint: Mulitply by
$$
\frac{\left(\sqrt[3]{n^2+5}\right)^2 + \sqrt[3]{n^2+5}\sqrt[3]{n^2+3} + \left(\sqrt[3]{n^2+3}\right)^2}{\left(\sqrt[3]{n^2+5}\right)^2 + \sqrt[3]{n^2+5}\sqrt[3]{n^2+3} + \left(\sqrt[3]{n^2+3}\right)^2}
$$
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

$$\lim\_{n\to\infty}(\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3})=\lim\_{n\to\infty}{2\over3c\_n^{2/3}}=0$$