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: 
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 $$\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}=\frac{2}{3}n^{-4/3}+O(n^{-10/3})$$

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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 $f'(n^2+2)\to 0.$
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

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