Suppose that f(x) = mx + b with m > 0. Let c be a real number. Prove that
lim x-> c (f(x)) = mc + b.
All that looks good so far. In looking at the definition of limit, you've got
$\displaystyle \displaystyle{\lim_{x\to c}f(x)=L}$ if for every $\displaystyle \epsilon>0$ there exists a $\displaystyle \delta>0$ such that if $\displaystyle 0<|x-c|<\delta$, then $\displaystyle |f(x)-L|<\epsilon.$
Overall, your goal is to find a $\displaystyle \delta(\epsilon)$ such that these conditions are satisfied. So, what do you want $\displaystyle \delta$ to be?
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.
Q.E.D.
I probably have too many of my initial calculations in there, but I want to make sure it flows from start to finish.