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Proof for {lim of exp = exp of lim}

  1. Jan 29, 2016 #1
    I don't know how to Google appropriately for this, since the kind of keywords I use present me with search results that try to define the exponential function using limits instead of what I am trying to ask:

    What does the proof look like for the following (assuming f(x) is "nice"). Any sites that can show this? Thanks.

    $$\lim_{x\rightarrow c} e^{f(x)}=e^{\displaystyle\lim_{x\rightarrow c} f(x)}$$
     
  2. jcsd
  3. Jan 29, 2016 #2

    blue_leaf77

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    Consider expanding the exponential into power series then use the distributive property of limit.
     
  4. Jan 29, 2016 #3

    fresh_42

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  5. Jan 29, 2016 #4

    dextercioby

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    That is a circular proof. In order to Taylor/MacLaurin expand a function, you need to have proven its continuity.
     
  6. Jan 30, 2016 #5

    blue_leaf77

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  7. Jan 30, 2016 #6

    Svein

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    Hm. I wonder...

    Define a function Φ(x) as Φ(x)=0 for x≤0; [itex] \phi(x)=e^{-\frac{1}{x^{2}}}[/itex] for x>0. Now Φ(x) is "nice" as it is C, even at x=0.
     
  8. Jan 30, 2016 #7

    fresh_42

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    Maybe, but is smoothness of ##f## necessary? Without considering details it looks like pointwise convergence at ##c## should be enough.
     
  9. Jan 30, 2016 #8
    It seems that it is just an application of the rule of composition of limits (if ##f## nice means that it has a finite limit in ##c##).

    The rule states that if ##h(x) \to b ## as ## x\to a##, and if ##g(x) \to \ell ## as ##x \to b##, then ##g(h(x)) \to \ell ## as ##x\to a##.
    If you add ##g## continuous (like ##x\to\exp(x)##) then ##\ell = g(b)##
     
    Last edited: Jan 30, 2016
  10. Jan 30, 2016 #9

    WWGD

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    I think , like Fresh_42 said, that continuity is one of the simplest ways of going about it, continuity implies sequential continuity (though the converse is not true)
     
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