<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] 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 $\lim (\sqrt[3]{n^2 + 5} - \sqrt[3]{n^2 + 3})=$
$\lim (\sqrt[3]{n^2 + 5} - \sqrt[3]{n^2 + 3})\frac {\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}{\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}=$
$\lim \frac{(n^2 + 5)-(n^2 +3)}{\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}=$
$\lim \frac{2}{\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}=0$
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