Proving lim(x->a) e^f(x) = e^lim(x->a)f(x): Struggling With Epsilon-Delta!

[forty]

I'm trying to prove that:

lim(x->a) e^f(x) = e^lim(x->a)f(x) (Assume lim(x->a)f(x) exists)

However I am having great difficulty! My only real approach I have taken is epsilon-delta proof.

if $$\epsilon$$ > 0 then there exists $$\delta$$ > 0 such that if |x - a| < $$\delta$$ then | e^f(x) - e^lim(x->a)f(x)| < $$\epsilon$$.

Assume $$\epsilon$$ > 0
Assume |x - a| < $$\delta$$

then to show that | e^f(x) - e^lim(x->a)f(x)| < $$\epsilon$$

Here I really have no idea...

As the limit of f(x) exits and that |x - a| < $$\delta$$ then by definition lim(x->a)f(x) = L (some number)

which gives

| e^f(x) - e^L| < $$\epsilon$$

Here I have no idea.. Do I use logs? Taylor series?

This has started to give me quite a headache! I've done proofs like this before but with concrete numbers not functions.

Any help on this as always would be greatly appreciated!

 

[Mark44]

Do you have to do this as an epsilon-delta proof? Do you have any theorems about, say, limits of continuous functions?

 

[hamster143]

It should be helpful to know that, if $$|f(x) - F| < \delta, |e^{f(x)} - e^F| <= |f(x)-F| e^F e^{\delta}$$.

 

[forty]

Mark44
I had a look at the properties and found nothing really useful. And I do think that I had to use an epsilon-delta proof >.< (they get very cumbersome very fast!)

hamster143
I don't know where that property comes from and I don't really have any idea how to apply it to the proof.

I should stick to physics.

 

[Mark44]

Why do you think you have to do an epsilon-delta proof? If the problem doesn't say that you have to prove it using delta and epsilon, then why assume that it does? What properties (actually theorems) do you have available? For instance, is there a theorem about the limit of a sum of functions? Product of functions? You said you found nothing useful, but maybe there was something there that could be used and you didn't recognize it.

 

[hamster143]

hamster143
I don't know where that property comes from and I don't really have any idea how to apply it to the proof.

I should stick to physics.

You're trying to do this as epsilon-delta proof. To do that, for any epsilon you need to find delta such that $$| e^{f(x)} - e^F | < \epsilon$$ for all |x - a| < $$\delta$$.

You know that f(x) converges to F, therefore you can find delta<1 such that |f(x)-F| < $$\epsilon / e^F e^1$$ for all |x - a| < $$\delta$$.

But if it is the case, using the property I showed, $$|e^{f(x)} - e^F| < \epsilon$$.

 

[forty]

Thanks for the help. Finally got to the bottom of it with help from a mate.

Mark44 you where right I didn't need to do an epsilon-delta proof, simple a LHS = RHS using limit properties and you hinted at! Just so use to seeing epsilon-delta proofs everywhere in real analysis.

But in short took the taylor series of one side and then used limit properties until it matches the other!

Thanks again!

 

[Mark44]

"If the only tool you have is a hammer, everything looks like a nail."