Thread: Proof - Limit of a function

Suppose that f(x) = mx + b with m > 0. Let c be a real number. Prove that
lim x-> c (f(x)) = mc + b.

What ideas have you had so far?

I know that |mx + b -(mc + b)| < \epsilon. So |mx + b -mc -b| < \epsilon. Then |mx -mc| < \epsilon. This can be factored as |m||x-c|<\epsilon. But I am not sure where to go from h ere.

All that looks good so far. In looking at the definition of limit, you've got

if for every there exists a such that if , then

Overall, your goal is to find a such that these conditions are satisfied. So, what do you want to be?

Ah, so if I let \delta = \epsilon/m, then |m(x-c)|<\epsilon which is what I am trying to prove. Right? Thank you!

Close. Note that So that requires what modification?

\epsilon/m > 0.

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.

Looks good to me. Your initial calculations in post # 3 are the ones that you have to reverse in the actual proof. So this is a very typical delta-epsilon proof.

I'd say you're done!