Epsilon-Delta Definition of Limits

  • Thread starter Thread starter lizielou09
  • Start date Start date
  • Tags Tags
    Definition Limits
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
lizielou09
Messages
3
Reaction score
0
Using the epsilon-delta definition of limit, prove that the limit as x approaches 0 of f(x) equals the limit as x approaches a of f(x-a).


I let the limit as x approaches 0 of f(x) equal A and the limit as x approaches a of f(x-a) equal B. If the absolute value of (A-B) is less than epsilon for all positive epsilon, then A=B. If A does not equal B, then let epsilon equal the absolute value of (A-B) divided by 2. Where do I go from here?
 
Physics news on Phys.org
Since you are give that limit as x --> 0 of f(x) = A you have the following statement:

Given ε >0 there is a δ > 0 such that |f(x) - A| < ε when 0 < |x| < δ

And you are trying to prove limit as x --> a of f(x-a) = A which is:

Given ε >0 there is a δ > 0 such that |f(x-a) - A| < ε when 0 < |x-a| < δ

If you think about it carefully you should be able to see how the first statement implies the second one.