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bsaucer
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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) (x^{p}-1)/p.
Example: f(z)= limit (as p approaches 0) (x^{p}-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.##bsaucer said:Is it possible for a limit of a range of functions to return a function?
Example: f(z)= limit (as p approaches 0) (x^{p}-1)/p.
With l'Hôspital's rule.bsaucer said:Assuming the parameter p is real, How do we arrive at the fact that the limit of the functions approaches the function Log x?
fresh_42 said:With l'Hôspital's rule.
\begin{align*}bsaucer said:Assuming the parameter p is real, How do we arrive at the fact that the limit of the functions approaches the function Log x?
I don't see how the equality follows here.julian said:\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*}
Just do ##e^{p \log x} = 1 + p \log x + \frac{1}{2!} (p \log x)^2 + \cdots##.WWGD said:I don't see how the equality follows here.
... and add an argument why the limits can be exchanged:julian said:Just do ##e^{p \log x} = 1 + p \log x + \frac{1}{2!} (p \log x)^2 + \cdots##.
I remember that result on exchanging limits named after one of the major names. Was it Lagrange, or Euler? Someone else?fresh_42 said: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 don't know. My source calls it the theorem of dominated convergence and uses Fatou's lemma to prove it.WWGD said:I remember that result on exchanging limits named after one of the major names. Was it Lagrange, or Euler? Someone else?
Ah, one of the Lebesgue convergence theorems; MCThm or DCT. Maybe I'm thinking of another resultfresh_42 said:I don't know. My source calls it the theorem of dominated convergence and uses Fatou's lemma to prove it.
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:WWGD said:Ah, one of the Lebesgue convergence theorems; MCThm or DCT. Maybe I'm thinking of another result
... and the Chrome translation trick doesn't work on pdf.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!)
bsaucer said:Is it possible for a limit of a range of functions to return a function?
Example: f(z)= limit (as p approaches 0) (x^{p}-1)/p.