Let $g(x)=\sqrt[3]{x}$ prove that g is continuous at c= 0 Please check my proof 
Let $\epsilon >0$ and $\delta >0$
$$|x-0|<\delta \leftrightarrow |\sqrt[3]{x}-\sqrt[3]{0}|<\epsilon $$
           $$  \leftrightarrow  \sqrt[3]{x}<\epsilon $$
           $$  \leftrightarrow  x<\epsilon ^{3}$$
choose $\delta =\epsilon ^{3}$
then 
$$|\sqrt[3]{x}-\sqrt[3]{0}|< \delta \leftrightarrow |\sqrt[3]{x}-\sqrt[3]{0}<(\sqrt[3]{\epsilon })^{3}=\epsilon $$
therefore it is continouos at 0
 A: You've done most the work correctly.  But some comments on your writeup:


*

*The first block of work between “Let $\epsilon > 0$” and “Choose $\delta = \epsilon^3$” is scratch work.  It shouldn't be included in your final product.

*You write “Let $\epsilon > 0$ and $\delta > 0$” at the start; this is not idiomatic.  You are fulfilling a definition that starts “For every $\epsilon > 0$, ...”  In other words, $\epsilon$ is arbitrary.  But $\delta$ is far from arbitrary; you have to specify it.  
So here is I would write it:

Given $\epsilon > 0$, let $\delta = \epsilon^3$.  Then for any real number $x$,
  $$
|x - 0 | < \delta \implies -\delta < x < \delta \implies -\epsilon^3 < x < \epsilon^3
$$
  Taking cube roots,
  $$
    -\epsilon < \sqrt[3]{x} < \epsilon
    \implies |\sqrt[3]{x} - 0 | < \epsilon
$$
  Since this is true for any $\epsilon$, we have $\lim_{x\to 0} \sqrt[3]{x} = 0$.  

This is probably how you are expected to write up the problem.  However, a question for your teacher: How do we know that
$$
    -\epsilon^3 < x < \epsilon^3
    \implies -\epsilon < \sqrt[3]{x} < \epsilon
$$
is true?  How, in fact, do we know what $\sqrt[3]{x}$ is when $x$ is a real number?  In fact, we don't know how to define $\sqrt[3]{x}$ until we know that the function $g(x) = x^3$ is continuous, and has a continuous inverse.  So this whole exercise puts the cart before the horse in a way.
