Let
Q:
a) Prove that g is continuous at c=0
b) Prove that g is continuous at a point . (The identity will be helpful.)
A: a) Let . Then, we have to show that, if it follows that
.
So, if , then implies . Therefore, 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.
Thanks
There's a little trick that helps in calculations like this. Let's agree from the start that we are going to choose (but we'll keep open the option of making futher restrictions on later on).
Then if , it follows that |x| must lie between 0 and 2|c|. In particular, . So . This tells us that .
Now we make a further restriction on , namely . That will ensure that .
Putting it all together, the way to choose is to require that .
In fact, my previous comment was completely misdirected. It showed how to prove the continuity of the function x^3, not x^{1/3}. So let's try again.
You want to show that x^{1/3} is continuous at x=c, where c≠0, and you have the identity . Apply the identity with and . That gives , from which . We can make the numerator of that fraction small, so we must ensure that the denominator does not also get too small. Therefore we want a lower bound on the size of the denominator.
Going back to the expression , if a and b have the same sign then all three terms in that expression are positive, and so . Therefore, if we can ensure that x and c have the same sign, it will follow that . But we can ensure that x and c have the same sign, by requiring that (because that tells us that |x-c|<|c|).
Thus if we know that . If we also knew that then it would follow that , as required.
So the way to choose is to require that .