Sorry, I still had some figuring to do on my own.
Clearly, for f=|x|, f is not differentiable at c=0.
However, fs(c)=lim(h->0)[f(c+h)-f(c-h)]/2h=0, as |x| is symmetric about the y-axis. That is, f(c+h)-f(c-h) is always zero.
Seems right to me. What do you think?
Big Help. Thank you snipez but, as the third part of the equality, did you mean to write f(c-h) instead of f(c+h)? As in
f'(c)=~=~=lim(h->0)[f(c)-f(c-h)]/h ?
I believe that's correct, and gives me what I need to proceed to part 1.
For part two I'm trying to think of a function that will...
Homework Statement
Def. f is Schwarz Differentiable at a pt c in its domain if
lim(h->0) [f(c+h)-f(c-h)]/2h exists as a finite limit.
1.)Prove or disprove: f is differentiable at c => f is Schwarz Differentiable at c
2.)Prove or disprove: f is Schwarz Differentiable at c => f is...