Prove that if f(x) -> A and g(x) -> B, then f(x)^g(x) -> A^B

  • Thread starter Ryker
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  • #1
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Homework Statement


Let [tex]\lim_{x\to c} f(x) = A; A > 0[/tex], and [tex]\lim_{x\to c} g(x) = B[/tex].

Prove that [tex]\lim_{x\to c} f(x)^{g(x)} = A^{B}[/tex].

The Attempt at a Solution


I really don't know how to approach this. I've been looking at it for an hour now, and while it seems obvious, I just don't know how to prove it. I've tried reducing it to limits of sequences and setting up an delta-epsilon proof, but it doesn't seem to pan out. Any ideas?
 

Answers and Replies

  • #2
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What about taking the log of both sides?
 
  • #3
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You mean log of the limit on one side and log of AB on the other? What base would I then take, A or f(x)? I've tried both, but still don't see where that helps to solve this. I mean, I believe this is the way it's done, I just don't see how.
 
  • #4
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The base of the log doesn't matter? The log just helps in changing the exponentiaton into a multiplication. And I suppose you've proven the result already for multiplication... (if not, then it's not that difficult).

So, what do you get if you take the log of both sides (say: log with base e. Like I said, the base doesn't matter...)
 
  • #5
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Alright, so I get [tex]\log \lim_{x\to c} f(x)^{g(x)} = B \log A[/tex].

I saw somewhere you can take log on the inside (why?), so that you get

[tex]\lim_{x\to c} (g(x) \log f(x)) = B \log A[/tex]
[tex]\lim_{x\to c} g(x) \lim_{x\to c} \log f(x) = B \log A[/tex]
[tex]\lim_{x\to c} \log f(x) = \log A[/tex]
[tex]\lim_{x\to c} (\log f(x) - \log A) = 0[/tex]
[tex]\lim_{x\to c} \log \frac{f(x)}{A} = 0[/tex]
[tex]\log \lim_{x\to c}\frac{f(x)}{A} = 0[/tex]

Since log1 = 0, the first equality should hold. Is this the way to do it then? And again, I'm having trouble with understanding why you can just take log in and out of the limit like that. I know it's continuity, but how exactly?

edit: For example, we don't even know if [tex]\lim_{x\to c} f(x)^{g(x)}[/tex] exists, we only know it's bounded.
 
Last edited:
  • #6
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It's because of following theorem:

If [tex]\lim_{x\rightarrow a}{f(x)}=b[/tex] and if g is continuous in b, then [tex]\lim_{x\rightarrow a}{g(f(x))}=g(b)[/tex].

This theorem (or something very much like this) HAS to be in your course somewhere! If it isn't, then your textbook is no good!! But it's also not very difficult to prove using epsilon-delta...
 
  • #7
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We don't have it stated explicitly like that, but there are some other propositions that are very similar and from which I should've been able to deduce that. Damn, I feel ridiculous now. I'm kind of interested how you'd prove it using an epsilon-delta proof, though, care to elaborate? The reason I'm asking this is because while we probably would've used such a proof somewhere along the way, as well, but not directly, so if there's a straight-forward way, which is different from ours, I'd like to see that.

And thanks for all your help thus far!
 
  • #8
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The epsilon-delta goes like this:

Se we need to prove

[tex]\forall \epsilon:~\exists \delta:~\forall x:~0<|x-a|<\delta~\Rightarrow |g(f(x))-g(b)|<\epsilon[/tex].

By continuity of g at b, we have that for a given [tex]\epsilon[/tex]:

[tex]\exists \delta^\prime:~\forall x:~|x-b|<\delta^\prime~\Rightarrow~|g(x)-g(b)|<\epsilon[/tex].

Since, [tex]\lim_{x\rightarrow a}{f(x)}=b[/tex], this implies that there exists a [tex]\delta[/tex], such that

[tex]0<|x-a|<\delta~\Rightarrow~|f(x)-b|<\delta^\prime[/tex].

Thus if, [tex]0<|x-a|<\delta[/tex], then [tex]|f(x)-b|<\delta^\prime[/tex], and thus [tex]|g(f(x))-g(b)|<\epsilon[/tex]. Which is exactly what we wanted to show!
 
  • #9
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Nice, thanks again :smile:
 
  • #10
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Hmm, alright, so for anyone that might be looking at this thread in the future, the proof described above for my original question is only good if you assume [tex]\lim_{x\to c} f(x)^{g(x)}[/tex] exists. But since usually the problem doesn't give you that assumption, the way to go about it is to express [tex]f(x)^{g(x)} = e^{\ln(f(x)^{g(x)})}[/tex] and then proceed from there, using the theorem micromass has stated above.
 

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