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Limits involving exponential functions

  1. Aug 14, 2009 #1
    1. The problem statement, all variables and given/known data

    I was studying L'Hopital's rule and how to deal with indeterminate forms of the type 0^0.

    It's not clear to me how lim e^f(x) = e^lim f(x).

    In wikipedia http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule" [Broken] (under other indeterminate forms)
    it says "It is valid to move the limit inside the exponential function because the exponential function is continuous".

    But that would mean lim(x->a) e^f(x) = e^f(a).

    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 14, 2009 #2

    rock.freak667

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    [tex]\lim_{x \rightarrow a}e^{f(x)} = e^{\lim_{x \rightarrow a} f(x)}[/tex]


    But for limx->af(x)=f(a) would only occur if f(x) is continuous at x=a
     
    Last edited: Aug 14, 2009
  4. Aug 14, 2009 #3
    Remember that f(x) may have issues at a, as mentioned above, even when it is part of e^f(x). You can move the limit inside the exponential, because the exponential itself doesn't have problem spots ("is continuous everywhere"), so it is only the f(x) inside that you have to deal with regarding the limit.
     
  5. Aug 14, 2009 #4
    Yea it makes sense, but is there some way to show this as a proof?
    For instance the basic limit properties such as lim(f + g) = lim f + lim g are proved.
     
  6. Aug 14, 2009 #5
    It should be straightforward to show from the limit definition of "continuous function" that if g is continous at L and [tex]\lim_{x\rightarrow a} f(x) = L[/tex], then [tex]\lim_{x\rightarrow a} g(f(x)) = g(L)[/tex].
     
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