Proof for {lim of exp = exp of lim}

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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.

$$\lim_{x\rightarrow c} e^{f(x)}=e^{\displaystyle\lim_{x\rightarrow c} f(x)}$$
 
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Consider expanding the exponential into power series then use the distributive property of limit.
 
blue_leaf77 said:
Consider expanding the exponential into power series then use the distributive property of limit.

That is a circular proof. In order to Taylor/MacLaurin expand a function, you need to have proven its continuity.
 
Hm. I wonder...

Define a function Φ(x) as Φ(x)=0 for x≤0; \phi(x)=e^{-\frac{1}{x^{2}}} for x>0. Now Φ(x) is "nice" as it is C, even at x=0.
 
blue_leaf77 said:
Well I interpret the OP's wording of "assuming f(x) is "nice"" as to be having all orders of derivative.
Maybe, but is smoothness of ##f## necessary? Without considering details it looks like pointwise convergence at ##c## should be enough.
 
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)##
 
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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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