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?
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.
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 $g(0)=1$
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 $f'(x) = 1/x$.
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 $e^{\delta \ln(x)} - 1 \underset{\delta \to 0}{\sim} \delta \ln(x)$
– Wirius Aug 22 '17 at 09:57This 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}
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 :)
