# Limit as a function, not a value

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bsaucer
Is it possible for a limit of a range of functions to return a function?
Example: f(z)= limit (as p approaches 0) (xp-1)/p.

Mentor
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Is it possible for a limit of a range of functions to return a function?
Example: f(z)= limit (as p approaches 0) (xp-1)/p.
Yes. You define ##f_p(x)=f(p,x)=\dfrac{x^p-1}{p}## and ask for ##\lim_{p \to 0}f(p,x)=\log x.##

Gold Member
Just consider the sequence of fuctions ##f_n(x)= x/n ##. Or a Taylor Series.

bsaucer
Assuming the parameter p is real, How do we arrive at the fact that the limit of the functions approaches the function Log x?

Mentor
2022 Award
Assuming the parameter p is real, How do we arrive at the fact that the limit of the functions approaches the function Log x?
With l'Hôspital's rule.

However, your question can only be answered if you first tell what the logarithm is for you and which tools are allowed in such a proof.

Homework Helper
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With l'Hôspital's rule.

Setting $f(p) = x^p$ we have by definition of the derivative $$\lim_{p \to 0} \frac{x^p - 1}{p} = \lim_{p \to 0} \frac{f(p) - f(0)}{p} = f'(0)$$ whenever the limit exists. There is no need to invoke l'Hopital's rule in such a case; it won't work for the example $$\lim_{x \to 0} \frac{g(x)}{x}$$ where $g(x) = x^2 \sin (x^{-1})$ for $x \neq 0$ with $g(0) = 0$, where $g'(0) = 0$ but $\lim_{x \to 0} g'(x)$ does not exist.

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Mentor
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I cited what Wikipedia said without checking, my bad. Here is the solution

\begin{align*}
\log x&=\int_1^x \dfrac{dt}{t}=\int_1^x\lim_{p\to 0}\dfrac{dt}{t^{1-p}}=\lim_{p\to 0} \int_1^x \dfrac{dt}{t^{1-p}}=\lim_{p\to 0} \left[\dfrac{t^p}{p}\right]_1^x=\lim_{p\to 0}\left(\dfrac{x^p}{p}-\dfrac{1}{p}\right)=\lim_{p\to 0}\dfrac{x^{p}-1}{p}
\end{align*}

Integral and limit can be exchanged because there exists an integrable majorant to ##f_n=t^{1-(1/n)}.##

Gold Member
Assuming the parameter p is real, How do we arrive at the fact that the limit of the functions approaches the function Log x?
\begin{align*}
\lim_{p \rightarrow 0} \dfrac{x^p - 1}{p} = \lim_{p \rightarrow 0} \dfrac{e^{p \log x} - 1}{p} = \log x .
\end{align*}

Gold Member
\begin{align*}
\lim_{p \rightarrow 0} \dfrac{x^p - 1}{p} = \lim_{p \rightarrow 0} \dfrac{e^{p \log x} - 1}{p} = \log x .
\end{align*}
I don't see how the equality follows here.

Gold Member
I don't see how the equality follows here.
Just do ##e^{p \log x} = 1 + p \log x + \frac{1}{2!} (p \log x)^2 + \cdots##.

WWGD
Mentor
2022 Award
Just do ##e^{p \log x} = 1 + p \log x + \frac{1}{2!} (p \log x)^2 + \cdots##.
... and add an argument why the limits can be exchanged:
$$\lim_{p\to 0}\lim_{n\to \infty }\sum_{k=0}^n \dfrac{(p\log x)^k}{k!}=\lim_{n\to \infty }\sum_{k=0}^n \lim_{p\to 0}\dfrac{(p\log x)^k}{k!}$$

Gold Member
I cited what Wikipedia said without checking, my bad. Here is the solution

\begin{align*}
\log x&=\int_1^x \dfrac{dt}{t}=\int_1^x\lim_{p\to 0}\dfrac{dt}{t^{1-p}}=\lim_{p\to 0} \int_1^x \dfrac{dt}{t^{1-p}}=\lim_{p\to 0} \left[\dfrac{t^p}{p}\right]_1^x=\lim_{p\to 0}\left(\dfrac{x^p}{p}-\dfrac{1}{p}\right)=\lim_{p\to 0}\dfrac{x^{p}-1}{p}
\end{align*}

Integral and limit can be exchanged because there exists an integrable majorant to ##f_n=t^{1-(1/n)}.##
I remember that result on exchanging limits named after one of the major names. Was it Lagrange, or Euler? Someone else?

Mentor
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I remember that result on exchanging limits named after one of the major names. Was it Lagrange, or Euler? Someone else?
I don't know. My source calls it the theorem of dominated convergence and uses Fatou's lemma to prove it.

Gold Member
I don't know. My source calls it the theorem of dominated convergence and uses Fatou's lemma to prove it.
Ah, one of the Lebesgue convergence theorems; MCThm or DCT. Maybe I'm thinking of another result

Mentor
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Ah, one of the Lebesgue convergence theorems; MCThm or DCT. Maybe I'm thinking of another result
Lebesgue is correct. It applies to any measurable function. I have found a nice PowerPoint presentation that covers most exchangeability results: sums <> limits <> integrals including counterexamples. Unfortunately, the only English part is a quotation at the beginning:
Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it before giving the problem to me.
(Richard Feynman, 1918–1988, Surely You’re Joking, Mr. Feynman!)
... and the Chrome translation trick doesn't work on pdf.
http://scratchpost.dreamhosters.com/math/HM3-D-2x2.pdf

WWGD