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