Limit as a function, not a value

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.

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.

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.##

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

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

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?

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.

fresh_42 said:

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.

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)}.##

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?

\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*}

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*}

I don't see how the equality follows here.

WWGD said:

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##.

julian said:

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:
$$
\lim_{p\to 0}\lim_{n\to \infty }\sum_{k=0}^n \dfrac{(p\log x)^k}{k!}=\lim_{n\to \infty }\sum_{k=0}^n \lim_{p\to 0}\dfrac{(p\log x)^k}{k!}
$$

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

WWGD said:

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.

fresh_42 said:

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

WWGD said:

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

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.

Just as a matter of terminology, a standard topic in mathematical analysis is whether a limit of a sequence of functions is another function. To study that question one must first define what "limit" means in the context of sequences of functions. The approach taken (implicitly) in this thread is to use the definition known as "pointwise convergence". There are other ways to define the limit of a sequence of functions. (See "uniform convergence").

The usual interpretation of a sequence of functions requires that it be a set of functions indexed by the integers. The set of functions of the form ##\frac{x^p -1}{p}## is a set indexed by the continuous parameter ##p##, so your question is technically more general than a question about a sequence of functions. In practical applications, I think most people would assume a set of functions indexed by a continuous parameter can be approximated by a set of functions indexed by the integers and they would pick a definition of "limit" from the list of standard definitions that apply to limits of sequences of functions.