# Proof for {lim of exp = exp of lim}

1. Jan 29, 2016

### DocZaius

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. Jan 29, 2016

### blue_leaf77

Consider expanding the exponential into power series then use the distributive property of limit.

3. Jan 29, 2016

### Staff: Mentor

4. Jan 29, 2016

### dextercioby

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

5. Jan 30, 2016

### blue_leaf77

6. Jan 30, 2016

### Svein

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.

7. Jan 30, 2016

### Staff: Mentor

Maybe, but is smoothness of $f$ necessary? Without considering details it looks like pointwise convergence at $c$ should be enough.

8. Jan 30, 2016

### geoffrey159

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
9. Jan 30, 2016

### WWGD

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)