Limits of e^f(x) as x Approaches Infinity

[Cherrybawls]

If I define the limit of f(x) as x aproaches infinity as "a", is it valid to say that the limit of
e^f(x) as x aproaches infinity is e^a?

 

[micromass]

Yes, this is correct.
To be 100% sure, I'll need to see the explicit limit...

 

[Cherrybawls]

The problem was:

lim x^(-1/x)
x->∞

 

[micromass]

Yes, this is correct!

 

[HallsofIvy]

$f(x)= e^x$ is continuous for all x so it is true that $\displaytype lim_{x\to\infty}e^{f(x)}= e^{\lim_{x\to\infty} f(x)}$.

Because you specifically ask about "$e^x$", I assume you mean that you are writing $x^x$ as $e^{ln(x^x)}= e^{x ln(x)}$.

 

[Cherrybawls]

Yes, HallsofIvy, that is essentially what I was doing, but it seems odd to me that limits behave in this way.

Is it also true then, that:
lim ln f(x)
x->∞
is the same as
ln lim f(x)
x->∞

 

[micromass]

Yes, this is also true. But you need to pay attention when the limit of f(x) is 0.

 

[Cherrybawls]

Is this a basic property of limits, or is there some theorem associated with this? Can you point me to some sort of reference or guide that could help me to better understand this?

 

[RadioactivMan]

It was taught to me as a theorem.

Let f and g be two functions such that:
1) lim g(x)=c
x->a
2) f is continuos in c

then

lim (f°g)(x)= lim x->a (f(g(x)))= f(lim x->a (g(x)))= f(c)
x->a

The proof is very similar to the one in page 146(spanish version) "Calculus" of Spivak. But the theorem in Spivak is a little more restrictive.
Actually that theorem is just a corollary of the one I wrote at the beginning.
The difference in the proof is that when Spivak applies the continuity of g, you have to apply the definition of the existence of the limit,then the rest is just the same.

 

[Cherrybawls]

Thank you so much, RadioactivMan, that was very helpful