Oh, good call. I didn't even think of trying anything by contradiction. I think that what you've written works perfectly!
(My previous post was in regard to my proof!)
Thanks!
Okay, so that proof doesn't work.
Here's why: when I'm given epsilon, I haven't actually chosen a value at which the function should be continuous. That is, I define the area that is continuous based on the choice of epsilon (and, therefore, delta). For this reason, the area that I'm...
Okay, so here's a solution that I just came up with (but I'm a bit iffy about using the limit to establish continuity near a point):
Lim(x->c+)[f] = L
=> Given epsilon > 0, there exists delta > 0 such that whenever 0<x-c<delta we have |f(x) - L| < epsilon/2.
Consider an arbitrary d such...
Thanks for you quick response!
Although this does provide the solution intuitively, I'm supposed to come up with a rigorous proof of this fact. Additionally, F actually is continuous at c (according to the fundamental theorem of calculus).
But, I agree, the area under the curve before c...
Homework Statement
Given: f is integrable on [a,b] and has a jump discontinuity at c in (a,b) (meaning that both one-sided limits exist as x approaches c from the right and from the left but that Lim(x->c-)[f] is not equal to Lim(x->c+)[f]).
Show that F(x) = Integrate[f,{a,x}] is not...