1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

[Spivak Calculus, Ch. 5 P. 9] Showing equality of two limits

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that ##\lim_{x \rightarrow a} f(x) = \lim_{h \rightarrow 0} f(a+h)##.

    2. Relevant equations

    By definition, if ##\lim_{x \rightarrow a} f(x) = l## then for every ##\epsilon > 0## there exists some ##\delta_1## such that for all x, if ##0<|x-a|<\delta_1## then ##|f(x)-l|<\epsilon##.

    Similarly, if ##\lim_{h \rightarrow 0} f(a+h) = m## then for every ##\epsilon > 0## there exists some ##\delta_2## such that for all h, if ##0<|h-0|<\delta_2## then ##|f(a+h)-m|<\epsilon##.

    3. The attempt at a solution

    I'm really not sure how to go from here. I think maybe I need to perform a proof by contradiction by assuming that ##l \neq m##, but I don't know what kind of contradiction I'm looking for.
     
    Last edited: Mar 30, 2013
  2. jcsd
  3. Mar 30, 2013 #2
    I think you need to prove that if the LHS has some limit l, then the RHS also has that same limit, and vice versa.
     
  4. Mar 30, 2013 #3
    Edit: actually, not quite sure what he's asking for here.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: [Spivak Calculus, Ch. 5 P. 9] Showing equality of two limits
Loading...