# Precise Def'n of Limit Proving the Sum Rule

1. Sep 13, 2009

Alrighty-then

I am going over a proof in Thomas' Calculus and I am not understanding a step.

Given that $\lim_{x\rightarrow c}f(x)=L \text{ and } \lim_{x\rightarrow c}g(x)=M$ prove that $\lim_{x\rightarrow c}(f(x)+g(x))=L+M$

Solution: He uses the triangle inequality to break up the summation

$|f(x)+g(x))-(L+M)|\le|f(x)-L|+|g(x)-M|$

Then he says:

"Since $\lim_{x\rightarrow c}f(x)=L$, there exists some number $\delta_1>0$ such that for all x

$0<|x-c|<\delta_1\Rightarrow|f(x)-L|<\epsilon/2$ "

I get lost there. Why $\epsilon/2$?

Thanks!

PS: Please don't yell at me Dick

2. Sep 13, 2009

### Dick

Because he is going to want to say that |f(x)-L|+|g(x)-M| is less than epsilon. If each term is less than epsilon/2 then they sum to less than epsilon. He's just picking the funny number 'e/2' so that he gets a nice neat 'e' at the end.

3. Sep 13, 2009

### slider142

No reason really. e/2 is an arbitrary number >0 just like e. He just plans to add another e/2 to it later to get e. He may just as well have started with e and the end of the proof would have 2e, which would have been fine as well, except for pedantry interested in getting the exact statement of a previous theorem/axiom.

4. Sep 13, 2009