I asked so many people and they don't understand it either. If you anybody could help it be much appreciated
Prove that f(x)=x^1/3 is continous at x=c for c>0
[Note: a³-b³=(a-b)(a²+ab+b²)]
It looks like he needs to do it by definition, ie. an $\displaystyle \epsilon - \delta$ proof.
@swtbellrose: Assume $\displaystyle |x-c| < \delta$ and use the hint: $\displaystyle \displaystyle |x-c| = \left| \left( x^{\frac{1}{3}} \right) ^3 - \left( c^{\frac{1}{3}} \right) ^3 \right| = \left| \left( x^{\frac{1}{3}} - c^{\frac{1}{3}} \right) \left(x^{\frac{2}{3}} + x^{\frac{1}{3}}c^{\frac{1}{3}} + c^{\frac{2}{3}} \right) \right|$
Now try to find a good $\displaystyle \delta$ so that $\displaystyle \left| x^{\frac{1}{3}} - c^{\frac{1}{3}} \right| < \epsilon$
But of course, since swtbellrose did not show any work at all, we have no idea what the problem really is. My first thought was just to note that $\displaystyle \lim_{x\to 0}x^{1/3}= 0$ which is the value of $\displaystyle 0^{1/3}$ so the function is continuous at x= 0.