Precise Def'n of Limit Proving the Sum Rule

  • #1
3,003
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Alrighty-then :smile:

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

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

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

[itex]|f(x)+g(x))-(L+M)|\le|f(x)-L|+|g(x)-M|[/itex]

Then he says:

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

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

I get lost there. Why [itex]\epsilon/2[/itex]?

Thanks!

PS: Please don't yell at me Dick :smile:
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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
1,015
70
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
3,003
6
Okay then. So it was just a 'choice' that he made. Not something that was 'dictated' by any of the given conditions.

That is, since we can get within 'e' of 'L' at some delta, then for the sake of the proof, let's go ahead and get within 'e/2' of 'L'
 

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