3
$\begingroup$

Consider

$$ f(x) = \lim_{\delta \to 0} \frac{x^\delta - 1}{\delta} $$

Then, naively, $$ \frac{df(x)}{dx} = \lim_{\delta \to 0} \frac{\frac{d{x^\delta}}{dx}}{\delta} = \lim_{\delta \to 0} \frac{\delta x^{\delta-1}}{\delta} = \lim_{\delta \to 0} x^{\delta-1} = \frac1{x}$$

what is $f$? And why? Are there any problems with using this?

$\endgroup$
  • $\begingroup$ You probably wanted to have $\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\lim_{\delta\to0}\frac{x^{\delta}-1}{\delta}=\lim_{\delta\to0}\frac{\frac{d}{dx}\left[x^{\delta}-1\right]}{\delta}=\lim_{\delta\to0}\frac{\delta x^{\delta-1}}{\delta}$, but I am not sure how you'd justify exchanging the limit and the derivative. $\endgroup$ – fiftyeight Aug 21 '17 at 11:13
  • $\begingroup$ @fiftyeight Whoops. THere was a typo. Fixed. The word "naively" covers the exchange of limits But what are rules for safely changing the order of limits? Are there none? $\endgroup$ – Peter Driscoll Aug 21 '17 at 11:27
  • $\begingroup$ I'd need to know more analysis to answer your question well. The question math.stackexchange.com/questions/409178/… seems related. $\endgroup$ – fiftyeight Aug 21 '17 at 12:13
  • $\begingroup$ I must say that I had never thought of exchanging the limit and derivative operation (as the limit could be easily evaluated using other techniques). +1 $\endgroup$ – Paramanand Singh Aug 21 '17 at 15:20
  • 1
    $\begingroup$ @ParamanandSingh Note that$$\frac{x^\delta-1}\delta=\int_1^xt^{\delta-1}~\mathrm dt$$So there is much intuition around this limit. $\endgroup$ – Simply Beautiful Art Aug 21 '17 at 23:55
7
$\begingroup$

The main problem with this approach is that you try to use the general argument

$$ \frac{\partial}{\partial y} \lim_{x \to a} f(x,y) = \lim_{x \to a} \frac{\partial}{\partial y} f(x, y) $$ but this is not a theorem. In more detail, what you need is

$$ \lim_{h \to 0} \frac{\lim_{x \to a} f(x, y+h) - \lim_{x \to a} f(x,y)}{h} = \lim_{x \to a} \lim_{h \to 0} \frac{f(x,y+h) - f(x,y)}{h} $$

and assuming the two limits in the numerator of the left hand side exist, this is equivalent to

$$ \lim_{h \to 0}\lim_{x \to a} \frac{f(x,y+h) - f(x,y)}{h}= \lim_{x \to a} \lim_{h \to 0} \frac{f(x,y+h) - f(x,y)}{h} $$

but this reverses the order of the limits — an operation that must be done with care, since it's not always true.

In order to make the argument work as is, you would need to find some way to justify why you can reverse the order of the limits.

A more typical way to continue is to use this argument as inspirational — now that you have a reasonable suspicion that $f'(x) = 1/x$, if you verify $f(1) = 0$ you should suspect that $f(x) = \ln x$, and you can search for an argument to verify that suspicion.

$\endgroup$
  • $\begingroup$ I think this touches on the core of the issue. There is a need to justify exchanging the limit and derivative $\endgroup$ – fiftyeight Aug 21 '17 at 11:16
  • $\begingroup$ Sorry. This answer is unsatisfactory to me as it does not explain how to get the limit or calculate the derivative. $\endgroup$ – Peter Driscoll Aug 21 '17 at 22:24
2
$\begingroup$

Let $g(t)=x^t$ and $x>0$

Then $$\ln{x}=g'(0)=\lim_{t \to 0} \frac{g(t) - g(0)}{t-0}=\lim_{t \to 0} \frac{x^t - 1}{t}$$

where $g(0)=1$

$\endgroup$
  • $\begingroup$ Confusing. You use delta outside the limits, as well as for the limit variable. Please clarify. $\endgroup$ – Peter Driscoll Aug 21 '17 at 22:28
  • $\begingroup$ So?..What is the definition of derivatives in a limit notion? $\endgroup$ – Marios Gretsas Aug 21 '17 at 22:30
  • $\begingroup$ $$f'(x_0)=\lim_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}$$ when $f$ is differentiable at $x_0$ of course $\endgroup$ – Marios Gretsas Aug 21 '17 at 22:31
  • $\begingroup$ @PeterDriscol..Now i used the same for the function $g(\delta)$ where $x=\delta $ and $x_0=0$ $\endgroup$ – Marios Gretsas Aug 21 '17 at 22:32
  • $\begingroup$ You mean $ g(t)=x^{t} $ ? So $g'(t) = \ln x x^t $ At $t = 0$ this is $ \ln x $. This works, but is confusing because of you use delta twice in two different contexts, the way you have written. So change the first use to t, and I am happy enough. Although it doesn't get to the heart of the matter. $\endgroup$ – Peter Driscoll Aug 21 '17 at 22:48
0
$\begingroup$

We have :

$$x^\delta - 1 = e^{\delta \ln(x)} -1 \underset{\delta \to 0}{\sim} \delta \ln(x)$$

By dividing by $\delta$, we obtain :

$$\frac{x^\delta - 1}{\delta} \underset{\delta \to 0}{\sim}\ln(x)$$

So indeed, $f = \ln$. From this follows that $f'(x) = 1/x$.

$\endgroup$
  • $\begingroup$ Where do you get $ e^{\delta \ln(x)} -1 \underset{\delta \to 0}{\sim} \delta \ln(x) $ $\endgroup$ – Peter Driscoll Aug 21 '17 at 22:30
  • $\begingroup$ To obtain this, one could use Taylor series to proove that. In fact, we have : $$e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}$$ This way, we have : $$e^{\delta \ln(x)} - 1 = \sum_{n = 1}^\infty \frac{(\delta \ln(x))^n}{n!} = \delta \ln(x) \underbrace{\sum_{n = 0}^\infty \frac{(\delta \ln(x))^n}{(n+1)!}}_{\underset{\delta \to 0}{\to} 1}$$ This prooves that $e^{\delta \ln(x)} - 1 \underset{\delta \to 0}{\sim} \delta \ln(x)$ $\endgroup$ – Wirius Aug 22 '17 at 9:57
0
$\begingroup$

Let me begin by saying that I am exploring an idea here. The idea I am exploring is how to have greater freedom in correctly using limit values. Please give me a little lee way.

Let me start informally choose a small enough p. For example, say $p = 2^{-64}$. The first equation can be expressed as.

$$ y = f(x) = \frac{x^p - 1}{p} $$ $$ (1 + yp)^{\frac1{p}} = x $$ Let $ yp = q $ then $p = \frac{q}{y}$ $$ ((1 + q)^{\frac1{q}})^y = x $$

Then the following equations hold, close enough, $$ e^y = x $$ $$ y = f(x) = \ln(x) $$

And I can make them hold as tightly as I want by getting out my calculator, and choosing a smaller p.

Now I want to make this a bit more formal. Let me define this in terms of X and Y being the limit values of x and y.

Let p be a small enough number so that x is close enough to a limit value X and y is close enough to a limit value Y. So I might choose an x = X and find a small enough p so that y is close enough to Y. Or I might choose a y = Y and find a small enough p so that x is close enough to X.

Let me try to make this follow a pattern like the $\epsilon$ $\delta$ limit defintition.

The limit as $p$ goes to zero of $x$ is $X$ and $y$ is $Y$ if for every $ \epsilon > 0 $ there exists a $ \delta $ such that, for all $p \in D$, if $ 0 < p < \delta $, then $ |x-X| < \epsilon \land |y-Y| < \epsilon$.

The equation for f is, $$ y = f(x) = \frac{x^p - 1}{p} $$ Let $ yp = q $ $$ ((1 + q)^{\frac1{q}})^y = x $$

Then, $$ e^Y = X $$ $$ Y = f(X) = \ln(X) $$

In this frame work, there is only a single p that goes towards zero. And there may be multiple limit values. $$ \frac{dy}{dx} = L $$ where $ |l - L| < \epsilon $ and, $$ l = \frac{f(x+p) - f(p)}{p} = \frac{((x+p)^p - 1) - (x^p - 1)}{p^2} = \frac{p^2 x^{p-1} + p^3 r}{p^2} = x^{p-1} + p r$$ Where r is other terms. So, $$\frac{dy}{dx} = L = x^{-1}$$

I hope this makes sense :)

$\endgroup$
  • $\begingroup$ This is more easily handled in the context of equations $$y=\left(1+\frac{x}{n}\right)^{n}\Leftrightarrow x=n(y^{1/n}-1)$$ which remains valid (surprise!) even when $n\to\infty$ leading to $$y=\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}\Leftrightarrow x=\lim_{n\to\infty} n(y^{1/n}-1)$$ $\endgroup$ – Paramanand Singh Aug 21 '17 at 15:27
  • $\begingroup$ This second bit, $y=\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}\Leftrightarrow x=\lim_{n\to\infty} n(y^{1/n}-1)$ has to be proved. $\endgroup$ – Peter Driscoll Aug 21 '17 at 15:36
  • $\begingroup$ Yes! But this comes as a surprise and it is not a consequence of the previous identity which is just basic algebra. Your manipulation assuming small $p$ is also similar nature. But I find the handling of the limits as $n\to\infty$ (in this case) somewhat easier than handling $p\to 0$. $\endgroup$ – Paramanand Singh Aug 21 '17 at 15:50
  • $\begingroup$ By the way, $y=e^x=(\left(1+\frac1{n}\right)^{n})^x)\Leftrightarrow x=n(y^{1/n}-1)$ is not correct for all n. $\endgroup$ – Peter Driscoll Aug 21 '17 at 15:54
  • $\begingroup$ Forgot to mention. $n$ is a positive integer greater than $|x|$. $\endgroup$ – Paramanand Singh Aug 21 '17 at 15:56
0
$\begingroup$

This is another way to deal with your problem of evaluation of the derivative of the limit in question.


The limit in question crucially depends on definition of symbol $x^{\delta} $. If the definition is based on exponential and logarithmic functions, then the answer $\log x$ is an immediate consequence of the properties of exponential and logarithmic functions.

If the definition of $x^{\delta} $ is independent of exponential and logarithmic functions then one can prove with some effort that the limit in question exists and defines a function, say $L(x) $ of $x$ for $x>0$. Further it can be proved that $$L(1)=0,L(xy)=L(x)+L(y),L'(x)=\frac{1}{x}$$ You may want to look at this answer for more details.


Assuming that the limit $$L(x) =\lim_{h\to 0}\frac{x^{h} -1}{h}$$ exists it is easy to see that $L(1)=0$ and as shown in linked answer we have $L(x) \leq x-1$. Next we have \begin{align} L(xy) &=\lim_{h\to 0}\frac{(xy)^{h}-1}{h}\notag\\ &=\lim_{h\to 0}x^{h}\cdot\frac{y^{h}-1}{h}+\frac{x^{h}-1}{h}\notag\\ &=1\cdot L(y) +L(x) \notag\\ &=L(x) +L(y) \notag \end{align}

Putting $y=1/x$ we get $L(1/x)=-L(x)$. We have further $$x-1\geq L(x) =-L(1/x)\geq - \left(\frac{1}{x}-1\right)=\frac{x-1}{x}$$ and thus $$\frac{1}{x}\leq \frac{L(x)}{x-1} \leq 1$$ for $x>1$. Letting $x\to 1^{+}$ and using Squeeze Theorem we get $$\lim_{x\to 1^{+}}\frac{L(x)}{x-1}=1$$ The limit above holds for $x\to 1^{-}$ also and it can be easily demonstrated by using substitution $x=1/t$. So we have finally arrived at $$\lim_{x\to 1}\frac{L(x)}{x-1}=1$$ or equivalently $$\lim_{x\to 0}\frac{L(1+x)}{x}=1$$ The calculation of derivative $L'(x)$ is now straightforward. We have \begin{align} L'(x) &=\lim_{h\to 0}\frac{L(x+h)-L(x)}{h}\notag\\ &=\lim_{h\to 0}\frac{L((x+h)/h)}{h}\notag\\ &=\lim_{h\to 0}\frac{L(1+(h/x))}{h/x}\cdot\frac{1}{x}\notag\\ &=1\cdot\frac{1}{x}\notag\\ &=\frac {1}{x}\notag \end{align}

$\endgroup$
  • $\begingroup$ $ x^\delta $ is defined as it is in maths. Normally as a power series, using a taylor expansion. $\endgroup$ – Peter Driscoll Aug 21 '17 at 22:36
  • $\begingroup$ @PeterDriscoll: Your comment does not provide any definition. It would be better if could explicitly write the definition you are using. $\endgroup$ – Paramanand Singh Aug 22 '17 at 1:17
  • $\begingroup$ By the way the definition of $x^{\delta} $ using a power series does seems rather unusual. If we proceed in this manner I think we have only two options : either assume binomial theorem for free or use exponential /logarithmic functions (this part is already mentioned in my answer). I wonder how useful the approach via binomial theorem would be. $\endgroup$ – Paramanand Singh Aug 22 '17 at 1:33
  • $\begingroup$ Probably a taylor series around $(x+1)^\delta$. $\endgroup$ – Peter Driscoll Aug 22 '17 at 2:07
  • $\begingroup$ @PeterDriscoll : so you assume binomial theorem for free. This works but I must say it is hardest definition (so far seen) to use in order to prove all the properties of the function. $\endgroup$ – Paramanand Singh Aug 22 '17 at 2:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.