It's known that lim, 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)| < \epsilon. => |\sqrt[3]{f(x)}| < \epsilon_0^3 < \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.
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