Suppose that f(x) = mx + b with m > 0. Let c be a real number. Prove that
lim x-> c (f(x)) = mc + b.
Oh, you're right. The problem assumed positive slopes. I was led astray by your unnecessary absolute values in post # 3. If the problem allowed negative slopes, you could just slap on absolute values and you'd be good to go.
So... a final proof looks like what?
Let \epsilon > 0. Choose \delta = \epsilon/m. Now if 0<|x-c|<\delta, then -\delta < x-c < \delta.
-\epsilon/m < x-c < \epsilon
-\epsilon < m(x-c) < \epsilon
|m(x-c)| < \epsilon
|mx - mc| < \epsilon
|mx + b - mc - b| < \epsilon
|mx + b - (mc + b)| < \epsilon
==> Lim x->c (mx + b) = mc + b.
I probably have too many of my initial calculations in there, but I want to make sure it flows from start to finish.