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Homework Help: Misunderstanding problem solution

  1. Dec 27, 2008 #1
    Hey all,

    1. The problem statement, all variables and given/known data

    This isn't really homework, just looking through a particular book (Spivak's Calculus, 3rd ed) and trying to see if my understanding of a problem is clear or not.

    Problem 9, page 107, for those of you following along at home.

    Prove that:

    [itex]\lim_{x \to a} f(x) = \lim_{h \to 0} f(a + h)[/itex]

    2. Relevant equations

    Now, the book gives the following answer:

    Let [itex]l = \lim_{x \to a} f(x)[/itex] and define [itex]g(h) = f(a + h)[/itex]. Then for every [itex]\epsilon > 0[/itex] there is a [itex]\delta > 0[/itex] such that, for all x, if [itex]0 < |x - a| < \delta[/itex] then [itex]|f(x) - l| < \epsilon[/itex]. Now, if [itex]0 < |h| < \delta[/itex] then [itex]0 < |(h + a) - a| < \delta[/itex] so [itex]|f(a + h) - l| < \epsilon[/itex]. This inequality can be written [itex]|g(h) - l| < \epsilon[/itex]. Thus [itex]\lim_{h \to 0} g(h) = l[/itex], which can be also written [itex]\lim_{h \to 0} f(a + h) = l[/itex]. The same sort of argument shows that if [itex]\lim_{h \to 0} f(a + h) = m[/itex], then [itex]\lim_{x \to a} f(x) = m[/itex]. So either limit exists if the other does, and in this case, they are equal.

    3. The attempt at a solution

    Now it appears to me that the author is making the a priori assumption that:

    [itex]\lim_{x \to a} f(x) = L[/itex] and [itex]\lim_{h \to 0} f(a + h) = L[/itex]

    and using that in his proof to show the two are equal. But I'm doubtful this is a legitimate jump.

    Let's do a different setup and I'll show you want I mean.

    Given [itex]\lim_{x \to a} f(x) = L[/itex] and [itex]\lim_{h \to 0} f(a + h) = M[/itex] , prove [itex]\lim_{x \to a} f(x) = \lim_{h \to 0} f(a + h)[/itex]

    [itex]0 < |x - a| < \delta[/itex] implies [itex]|f(x) - L| < \epsilon[/itex]

    Now if

    [itex]0 < |h| < \delta[/itex] then

    [itex]0 < |(h + a) - a| < \delta[/itex] implies [itex]|f(a + h) - M| < \epsilon[/itex]

    Oh wait, we can't go any further.. The two are not necessarily equal because L might not equal M.

    So I'm pretty sure the book is right and I'm wrong.. so what's the error in my logic?

    - Rob
     
  2. jcsd
  3. Dec 27, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The reason why |f(a+h)-L|<epsilon isn't because of a separate assumption that f(a+h) has a limit. It's because you can substitute h+a for x in the first limit. h+a is one of the values of x such that |x-a|<delta.
     
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