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Suppose

  • $\lim_{x \to a} f(x) = L$
  • $\lim_{x \to a} g(x) = M$

I would like to prove

$$\lim_{x \to a} f(x)^{g(x)} = \left(\lim_{x \to a} f(x)\right)^{\lim_{x\to a} g(x)}=L^M$$

I thought this had probably been asked before, but the closest I can find is Prove $\lim_{x \to a} f(x)^n = \left(\lim_{x \to a} f(x)\right)^n$.

All limit law proofs I found on Google excluded this one. However it seems to be true, as it is listed on this Wikipedia article (sadly without proof).

I imagine the proof is quite difficult, since $g$ can take on integer, rational, and irrational values in general, all of which would change the meaning of $f^g$.

Thanks in advance.

EthanAlvaree
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    You'd need conditions for this to work out, for example having zero/negative base and exponent wouldn't work out. Having said that, note that $\log(f(x)^{g(x)}) = g(x)\log f(x)$, so you can probably deduce any limit theorems as corollaries of the various theorems surrounding continuity (composition and product), and the continuity of the logarithm and exponential. – Sarvesh Ravichandran Iyer Jul 25 '21 at 10:14
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    I don’t think this is that hard to prove, although that’s because I happened to be thinking about this question yesterday. What have you tried? The proof should go along these lines: For $\varepsilon$, no matter how small, there exists $x$ is close enough to $a$ such that $f(x)$ is close enough to $L$ and $g(x)$ is close enough to $M$ such that $f(x)^{g(x)}$ is close to $L^M$. I think you can use Binomial expansion or Bernoulli’s Inequality or something to fill in the details… – Adam Rubinson Jul 25 '21 at 10:14
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    In fact, this question has been asked before. See here and here. Found using Approach0 like so. – Sarvesh Ravichandran Iyer Jul 25 '21 at 10:18

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