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**Danneedshelp** Let $\displaystyle g(x)=x^{1/3}$

Q:

a) Prove that g is continuous at c=0

b) Prove that g is continuous at a point $\displaystyle c\neq\\0$. (The identity $\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$ will be helpful.)

A: a) Let $\displaystyle \epsilon>0$. Then, we have to show that, if $\displaystyle |x-c|=|x-0|=|x|<\delta$ it follows that

$\displaystyle |g(x)-c|=|x^{1/3}-0|=|x^{1/3}|<\epsilon$.

So, if $\displaystyle \delta=\epsilon$, then $\displaystyle |x|<\delta$ implies $\displaystyle |x^{1/3}|<\epsilon$. Therefore, $\displaystyle g(x)$ is ontinuous at 0.

b) Given I am not sure about the first part, I am pretty stuck on the (b). I am not sure how to use the identity in my scratch work; I keep hitting a brick wall. Some help would be great.