Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$ It's known that $\lim\limits_{x \rightarrow x_0}f(x) = A$, how to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$?

Here's what I've got now:

When $A = 0$, to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = 0$: Since we have $\lim\limits_{x \rightarrow x_0}f(x) = A = 0$, so $|f(x)| &lt; \epsilon$. => $|\sqrt[3]{f(x)}| &lt; \epsilon_0^3 &lt; \epsilon$

When $A \ne 0$, $|\sqrt[3]{f(x)} - \sqrt[3]{A}| = \frac{|f(x) - A|}{|f(x)^{\frac{2}{3}}+(f(x)A)^{\frac{1}{3}} + A^{\frac{2}{3}}|}$...

How can I deal with $(f(x)A)^{\frac{1}{3}}$? Thanks.
You might want to take a look at the limit of composite functions. This is a standard result in many calculus textbooks. Fishing around online immediate gives several proofs, for example here or here.
You need to bound the term

$$ \frac{1}{|f(x)^{\frac{2}{3}}+(f(x)A)^{\frac{1}{3}} + A^{\frac{2}{3}}|}\,, $$

and exploit the fact that $|x-x_0|&lt;\delta \implies |f(x)-A|&lt;\epsilon\,.$ See here for details.