- #1

bsaucer

- 26

- 0

Example: f(z)= limit (as p approaches 0) (x

^{p}-1)/p.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter bsaucer
- Start date

- #1

bsaucer

- 26

- 0

Example: f(z)= limit (as p approaches 0) (x

- #2

- 17,826

- 19,064

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.##

Example: f(z)= limit (as p approaches 0) (x^{p}-1)/p.

- #3

WWGD

Science Advisor

Gold Member

- 6,336

- 8,392

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

- #4

bsaucer

- 26

- 0

- #5

- 17,826

- 19,064

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.

- #6

pasmith

Homework Helper

2022 Award

- 2,593

- 1,196

With l'Hôspital's rule.

Setting [itex]f(p) = x^p[/itex] we have by definition of the derivative [tex]\lim_{p \to 0} \frac{x^p - 1}{p} = \lim_{p \to 0} \frac{f(p) - f(0)}{p} = f'(0)[/tex] 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 [tex]

\lim_{x \to 0} \frac{g(x)}{x}[/tex] where [itex]g(x) = x^2 \sin (x^{-1})[/itex] for [itex]x \neq 0[/itex] with [itex]g(0) = 0[/itex], where [itex]g'(0) = 0[/itex] but [itex] \lim_{x \to 0} g'(x)[/itex] does not exist.

Last edited:

- #7

- 17,826

- 19,064

\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)}.##

- #8

julian

Gold Member

- 752

- 240

\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*}

- #9

WWGD

Science Advisor

Gold Member

- 6,336

- 8,392

I don't see how the equality follows here.\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*}

- #10

julian

Gold Member

- 752

- 240

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

- #11

- 17,826

- 19,064

... and add an argument why the limits can be exchanged:Just do ##e^{p \log x} = 1 + p \log x + \frac{1}{2!} (p \log x)^2 + \cdots##.

$$

\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!}

$$

- #12

WWGD

Science Advisor

Gold Member

- 6,336

- 8,392

I remember that result on exchanging limits named after one of the major names. Was it Lagrange, or Euler? Someone else?

\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)}.##

- #13

- 17,826

- 19,064

I don't know. My source calls itI remember that result on exchanging limits named after one of the major names. Was it Lagrange, or Euler? Someone else?

- #14

WWGD

Science Advisor

Gold Member

- 6,336

- 8,392

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

- #15

- 17,826

- 19,064

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: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!)

http://scratchpost.dreamhosters.com/math/HM3-D-2x2.pdf

Share:

- Last Post

- Replies
- 16

- Views
- 543

- Last Post

- Replies
- 3

- Views
- 465

- Replies
- 1

- Views
- 321

- Last Post

- Replies
- 29

- Views
- 906

- Replies
- 3

- Views
- 388

- Last Post

- Replies
- 3

- Views
- 430

- Last Post

- Replies
- 2

- Views
- 705

- Last Post

- Replies
- 4

- Views
- 582

- Last Post

- Replies
- 1

- Views
- 473

- Last Post

- Replies
- 16

- Views
- 830