I noticed that if I had a function $f(x)=x^n$ where $n$ is an integer, then $\lim_{m\to{n^+}}f^{(m)}(x)=n!$ where $f^{(m)}(x)$ is the $m$-th derivative. Also, $$\lim_{m\to{n^-}}f^{(m)}(x)=\frac{(-1)^n\ln(x)}{n!}$$Where we have $m,n$ as integer numbers.
The limits I use in this question are not limits of the normal sense. Instead, they are used to describe positives, negatives, and where it is heading.
$\lim_{m\to n^+}f^{(m)}(x)$ means that we are taking the $m^{th}$ derivative of $f(x)$ where we see $n$ is positive.
$\lim_{m\to n^-}f^{(m)}(x)$ means that we are taking the $m^{th}$ anti-derivative (it is negative) where we see $n$ is negative.
There is also a bridge between negative and positive $n$'s when $n$ is not an integer.
Which I though was pretty cool.
So I drew a little picture of what this was with an open circle in the middle for an unknown value at $\lim_{m\to{n}}f^{(m)}(x)$.
Basically, a number line for the powers of polynomials after the $n$-th derivative. But I noticed something peculiar.
The integral of $\ln(x)$ looks a lot like $x^n$. In fact, I deduced the following.
Let $g(x)=\ln(x)$$$g^{(-n)}(x)=\frac{x^n[\ln(x)-\sum_{i=1}^n\frac1i]}{n!}$$where the $-n$-th derivative is the $n$-th anti-derivative.
So I got curious and wondered that upon taking derivatives of such a function, I would return back to polynomials in the form $x^n$, but what if I approached this from the other side? Like we did with polynomials.
We could try to find a sort of bridge, where $n$ was non-integer like we did before. Then we could try to approach our original point which I will call $g_0(x)$ from the sort of negative side, in a sense that it is opposite of the other side.
My assumption is that it would result in something along the lines of $h(x)\approx x^n\ln^2(x)$ with a lot of extra constants.
But I have no way of proving this because my formula for the $n^{th}$ anti-derivative does not work for negative numbers.
So I hope someone can find this interesting value.
Lastly, if this is so true, then I wondered the opposite. Why would taking the derivative of a polynomial eventually reach $0$? If my assumptions are true, the derivatives of a polynomial from the positive side should reach $e(x)=x^ne^x$ (hehe, get it? e comes before f?)
EDIT: It would actually result in something along the lines of $\frac{x^n}{\ln(x)}$
Also, if you can repeat my pattern and find some $h(x)$, then what about an $i(x)$? Possibly a pattern I can use?
December $25^{th}$
I've also noted that the growth rates of $x^n\ln(x)$ exceed the growth rates of $x^n$.
If my assumptions are correct, then the growth rate keeps increasing.
Again, December $25^{th}$
I've also noticed that the growth rates of $x^n\ln(x)$ are less than the growth rates of $x^{n+a}$, meaning it is a sort of in-between function as far as growth rates. Is there anything we can conclude on where $x^n\ln(x)$ could exist on a number line with $x^{+n}$ on the left and $x^{-n}$ on the right?
I also can't find what this value $a$ is.
It is clear $x^a\ne\ln(x)$, but for $0<a<1$, we have almost logarithmic growth rates.
Actually, I have just realized that $a=1$.
That is: $$\lim_{a\to1^-}x^{n+a}<<x^n\ln(x)<<x^{n+1}$$ If I am using my symbols correctly.
What does this mean?
What other interesting properties can we deduce?
Part 1
I will start with your first paragraph. You misunderstood the notation of limits. First of all, you don't need limits to express what you mean.
Write down the derivatives of $x^n$ for $n\geqslant0$:
$$\underbrace{x^n}_{0-\text{th}}, nx^{n-1},n(n-1)x^{n-1},...,\underbrace{n!}_{n-\text{th}},...$$
We conclude that the general form of this is
$$\frac{d^k}{dx^k}x^n=\frac{n!}{(n-k)!}x^{n-k}$$ it implies that $$\frac{d^n}{dx^n}x^n=n!$$
Now consider the case for $n<0$. Of course we need then negative derivatives, exactly as you defined. To make things simpler we do a substitution $m=-n$ to have $m>0$. Then we need a general form of $$\int^k\frac{1}{x^m}\,dx^k$$ where $$\int^k\cdot \;dx^k=\underbrace{\iint\cdots\int}_{k\text{ times}}\cdot\;\underbrace{dx\,dx...dx}_{k\text{ times}}$$ means $k$-th integral (we will skip constants for simplicity). Then we write down few integrals
$$\frac{1}{x^m},\frac{-1}{(m-1)x^{m-1}},\frac{1}{(m-1)(m-2)x^{m-2}},...,\frac{(-1)^{m+1}}{(m-1)!x},\frac{(-1)^{m+1}}{(m-1)!}\ln(x)$$
it follows that $$\int^m\frac1{x^m}\,dx^m=\frac{(-1)^{m+1}}{(m-1)!}\ln(x)$$
We can here define a function $$f(x,n)=x^n$$
and write down our two relations as $$F(n)=\frac{\partial^n}{\partial x^n}f(x,n)=\begin{cases} \displaystyle x\mapsto n!&\text{for } n\geqslant 0\\[0.7em] \displaystyle x\mapsto\frac{(-1)^{n+1}}{(-n-1)!}\ln(x) & \text{for } n< 0 \end{cases}\tag1$$
Of course this is function $\mathbb Z\to\mathcal{C}^\infty(\mathbb R_+)$ but using analysis we are able to expand it to some bigger set. First note that we can expand $n!$ using $\Gamma(n+1)$ (by the way it is the only holomorphic logarithmically convex function satisfying $f(n+1)=n!$). The problem is when facing the $(-1)^{n+1}$ term. The only thing we can do is use the formula $$(-1)^z=e^{\pi i z}=\cos(\pi z)+i\sin(\pi z)$$ so the function $(1)$ looks like
$$F(n)=\begin{cases} \displaystyle x\mapsto\Gamma(n+1)&\text{for } n\geqslant 0\\[0.7em] \displaystyle x\mapsto\frac{-\cos(\pi n)-i\sin(\pi n)}{\Gamma(-n)}\ln(x) & \text{for } n< 0 \end{cases}$$
It turns out that this function is not continuous1 at $n=0$. Function $n!$ tends to $x\mapsto1$ whereas the second one tends to $x\mapsto0$:
$$\lim_{n\to0^-} F(n)=x\mapsto0 \quad\ne\quad x\mapsto1=\lim_{n\to0^+}F(n)$$
You can see the graph here (move the slider $n$ - this is the argument, plot is the value). Keep in mind that I've deleted the imaginary part (by setting $i=0$) to make this visible, but it tends to $0$ anyway.
Part 2
Your formula is correct:
$$\int^n\ln(x)\,dx^n=\frac{x^n(\ln(x)-H_n)}{n!}\tag2$$
Nevertheless no element of this family of functions is close to $x^k$, you can see the plot here.
But you're right. We are able to get back differentiating $\ln(x)$ $n$ times $$\frac{(-1)^{n+1}}{(n-1)!}\ln(x)\quad\xrightarrow{\text{diferentiate }n\text{ times}}\quad\frac{1}{x^n}\tag3$$
See the pattern for $n$-th derivative of $\ln(x)$:
$$\ln(x),\frac1x,-\frac1{x^2},\frac{2}{x^3},-\frac{3}{x^4},...$$
The general form looks like the following
$$\frac{d^n}{dx^n}\ln(x)=(-1)^{n+1}\frac{(n-1)!}{x^n}$$
And plugging into $(3)$ we get
$$\frac{d^n}{dx^n}\,\frac{(-1)^{n+1}}{(n-1)!}\ln(x)= \frac{{(-1)^{n+1}}}{{(n-1)!}}\cdot {(-1)^{n+1}}\frac{{(n-1)!}}{x^n}$$
Part 3
In part 2 I said that $(2)$ is not close to $x^k$. One may consider why is that if the limit
$$\lim_{n\to\infty}\left(\ln(n)-H_n\right)$$
converges to a constant. This constant is equal to $-\gamma$ where $\gamma$ is the Euler–Mascheroni constant. Of course it is because $$\ln(n)-H_n\not\equiv\ln(x)-H_n$$
Here is the plot of $(2)$ and $-\gamma x^n/n!$. A different thing is when you consider a function $$\frac{x^n(\ln(\color{red}{n})-H_\color{red}{n})}{n!}$$
it converges to $-\gamma x^n/n!$ as shown here.
Part 4
You said that $$\lim_{a\to1^-}x^{n+a}\ll x^n\ln(x)\ll x^{n+1}$$
whereas I am pretty sure you meant $$x^{n+a}\prec x^n\ln(x)\prec x^{n+1}\tag4$$
where $\alpha\prec\beta\iff\alpha\in o(\beta)$. The notation $\gg$ and $\ll$ is used in approximations and physics and means "much larger/smaller". To check whether $x^{n+1}$ grows faster than $x^n\ln(x)$, consider a limit
$$\lim_{x\to\infty} \frac{x^{n+1}}{x^n\ln(x)}=\infty$$
Indeed, we have $x^{n+1}\succ x^n\ln(x)$. Now the general form
$$L=\lim_{x\to\infty} \frac{x^{n+a}}{x^n\ln(x)}=\lim_{x\to\infty} \frac{x^a}{\ln(x)}$$
When $a=0$ we have of course $L=0$, but when $a>0$ we can use l'Hospital's rule to get
$$L=\lim_{x\to\infty} \frac{ax^{a-1}}{1/x}=\lim_{x\to\infty} ax^a=\infty$$
It means that only $a=0$ satisfies $(4)$.
1 By $F:\mathbb Z\to\mathcal C$ being continuous I mean that for $a,b$ such that $|a-b|$ is infinitesimal, the function $|F(a)-F(b)|$ is bounded by an infinitesimal.