# Thread: Can someone help me with this problem on continuity? :(

1. ## Can someone help me with this problem on continuity? :(

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)]

2. All you have to do is to show that the function is differentiable in the domain x>0 because a function must be continuous at a point in order for it to be differentiable at that point. Hope that helps!

3. Originally Posted by Linnus All you have to do is to show that the function is differentiable in the domain x>0 because a function must be continuous at a point in order for it to be differentiable at that point. Hope that helps!
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 $\displaystyle \left| x^{\frac{1}{3}} - c^{\frac{1}{3}} \right| < \epsilon$

4. 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.

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