Just instantiate epsilon to |h(x0)| / 2 in the definition of continuity of h at x0. What do you get?
Hi,
I have a question which I'm having trouble with.
I am told that I should first use the epsilon delta definition of continuity to prove that there must be a closed interval [c,d] with c < x0 < d such that for some x in [c, d], |h(x)| > |h(x0)|/2 by choosing epsilon to be |h(x0)|/2. I am having trouble with this.
Any help would be appreciated
Which, in turn, means that -h(x0)/2< h(x)- h(x0)< h(x0)/2. (You don't need absolute value on h(x0) because they are by hypothesis positive) and the absolute values on h(x) and h(x0) on the left are wrong.)
Now add h(x0) to each part. h(x0)- h(x0)/2= h(x0)/2< h(x)< h(x0)/2+ h(x0)= 3h(x0)/2. And, again, by hypothesis h(x0) is positive.