# Proof - Limit of a function

• Aug 25th 2010, 09:18 AM
tarheelborn
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.
• Aug 25th 2010, 09:21 AM
Ackbeet
What ideas have you had so far?
• Aug 25th 2010, 09:36 AM
tarheelborn
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.
• Aug 25th 2010, 09:57 AM
Ackbeet
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?
• Aug 25th 2010, 10:53 AM
tarheelborn
Ah, so if I let \delta = \epsilon/m, then |m(x-c)|<\epsilon which is what I am trying to prove. Right? Thank you!
• Aug 25th 2010, 10:54 AM
Ackbeet
Close. Note that $\displaystyle \delta>0.$ So that requires what modification?
• Aug 25th 2010, 11:08 AM
tarheelborn
\epsilon/m > 0.
• Aug 25th 2010, 11:51 AM
Ackbeet
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?
• Aug 25th 2010, 11:59 AM
tarheelborn
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.
• Aug 25th 2010, 12:01 PM
Ackbeet
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!