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Real analysis epsilon delta problem

  1. May 23, 2013 #1
    I've been reading through Spivak's calculus, and the problem is the answer key i have a hold of is for a different edition so it often doesn't answer the correct questions.

    Anyways, here they are:

    Chapter 5 problem 10

    b. Prove that lim x-> 0 f(x) = lim x-> a f(x-a)
    c. Prove that lim x-> 0 f(x) = lim x-> 0 f(x^3)

    I'm mainly having a problem with making my solution rigorous.

    For b what I have is on the left side we know there is a α_1 for every ε so

    |x| < α_1 and |f(x)-m| < ε

    We can choose the same epsilon for the right side:

    |x-a| < α_2 and |f(x-a)-n| < ε
    So the two x's are necessarily related so I replace it in the second equation with y and I also choose α=min(α_1,α_2).

    So we have:

    |x| < α and |f(x)-m| < ε
    |y-a| < α and |f(y-a)-n| < ε

    So now I want to say something like, let y-a=x so if ε<|m-n|/2 then it is impossible for both to be fulfilled unless m=n but I'm not sure how to say that rigorously.

    I have a similar problem with the next problem.
     
  2. jcsd
  3. May 24, 2013 #2

    Office_Shredder

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    You're making things too complicated at this step. For all [itex] \epsilon >0[/itex], there is a [itex] \delta[/itex] such that if [itex]|x|<\delta[/itex], then [itex] |f(x)-m|<\epsilon [/itex] for some m.

    Now, if [itex] |x-a|<\delta[/itex], the above sentence immediately says that [itex] |f(x-a)-m|<\epsilon[/itex] with no additional work
     
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