How can I prove that this limit is equal to the Natural Logarithm? Earlier, I was messing around a little bit with the reverse power rule/power rule for integration which is typically written as follows.
Let $f\colon\mathbb{R} \to \mathbb{R}$ be a function satisfying $f(x) = x^r$ for all $x$, with $r \in \mathbb{R}$. Then,
$$ \int f(x) \,\mathrm{d}x = \int x^r \,\mathrm{d}x = \frac{x^{r+1}}{r+1} + C $$
for any real number $r\neq -1$ and some arbitrary constant $C$.
I was trying to figure out what happens as $r\to -1$ and how in the limit it might approach the natural logarithm of $x$ because of
$$ \int x^{-1} \,\mathrm{d}x = \int \frac{1}{x} \,\mathrm{d}x = \begin{cases} \ln(x) + C &\text{if $x>0$} \\ \ln(-x) + C&\text{if $x<0$} \end{cases} = \ln|x| + C.$$
I am aware that
$$ \int_{1}^{x} t^{-1} \,\mathrm{d}t = \int_{1}^{x} \frac{1}{t} \,\mathrm{d}t = \ln(x) - \ln(1) = \ln(x) $$
and that
$$ \begin{aligned}
\lim_{r\to -1}\,\int_{1}^{x} t^{r} \,\mathrm{d}t &= \lim_{r\to -1} \left(\frac{x^{r+1}}{r+1} - \frac{1}{r+1} \right) \\[10pt]
&= \lim_{r\to -1} \frac{x^{r+1}-1}{r+1} \\[10pt]
&\overset{\scriptscriptstyle\text{L'H}}{=} \lim_{r\to -1} \frac{\frac{\mathrm{d}}{\mathrm{d}r} (x^{r+1}-1)}{\frac{\mathrm{d}}{\mathrm{d}r} (r+1)} \\[10pt]
&= \lim_{r\to -1} \frac{x^{r+1}\ln(x) \cdot 1}{1} \\[10pt]
&= \ln(x).
\end{aligned} $$
I tried graphing both $\dfrac{x^{r+1}}{r+1}$ and $\ln(x)$ on Desmos and after manually tinkering with the values of $r$ close to $-1$ (e.g., $-0.9,-0.99,-0.999$, etc.) for a little bit, I came up with this function $g\colon\mathbb{R} \to \mathbb{R}$ with $r \in \mathbb{R}$ and $n \in \mathbb{Z}^+$ where
$$ \begin{align}
r &= -\sum_{k=0}^{n-1}\, 9 \cdot 10^{k-n}, \tag{$\lim_{n\to\infty} r = -1$} \\[10pt]
g(x) &= \frac{x^{r+1}}{r+1} - 10^n.
\end{align} $$
From what I gathered, it seems that $g$ approaches $\ln(x)$ as $n$ increases.
Finally, my question is (if my assumption is true) how can I prove that the following limit is equal to the natural logarithm?
$$ \begin{align}
\lim_{n\to\infty} g(x) &= \lim_{n\to\infty} \left(\frac{x^{r+1}}{r+1} - 10^n\right) \\[10pt]
&= \lim_{n\to\infty} \left(\frac{x^{\displaystyle\left(-\sum_{k=0}^{n-1}\, 9 \cdot 10^{k-n}\right)+1}}{\displaystyle\left(-\sum_{k=0}^{n-1}\, 9 \cdot 10^{k-n}\right)+1} - 10^n \right) \\[10pt]
&\overset{\text{?}}{=} \ln(x). 
\end{align}$$
Thank you.
 A: Observe that the decimal number is essentially $1-10^{-n}$ so the limit of $g(x)$ is just the limit of $$ \dfrac{x^{-(1-10^{-n})-1}}{-(1-10^{-n})-1} - 10^n $$
which simplifies to $$ (x^{10^{-n}}-1)/10^{-n} $$ which is a standard limit
A: Okay so I've come up with a full solution after a while (also thanks to Aditya's suggestion)
$$\begin{align}
e :\!\!&= \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n \\[5pt]
e &= \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^{\frac{n}{x}} \\[7pt]
\exp(x) :\!\!&= e^x = \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^{n} \\[7pt]
\ln(x) :\!\!&= \text{inverse of } \exp(x) \tag{$f(f^{-1}(x))=f^{-1}(f(x))=x$} \\[7pt]
\exp(\ln(x)) &= \lim_{n\to\infty} \left(1 + \frac{\ln(x)}{n}\right)^{n} = x = \lim_{n\to\infty} x \\[7pt]
\ln(x) &= \lim_{n\to\infty} n(x^{\frac{1}{n}}-1)
\end{align}$$
Now we can show that
$$\begin{align}
r &= -\sum_{k=0}^{n-1}\, 9 \cdot 10^{k-n} = -\frac{1}{10^n} \sum_{k=0}^{n-1}\, 9 \cdot 10^{k} \tag{$\lim_{n\to\infty} r = -1$} \\[10pt]
&= -\left(\frac{9 + 90 + 900 + \cdots + 9\cdot 10^{n-1}}{10^n}\right) \\[10pt]
&= -\left[\frac{(10-1) + (100-10) + (1000-100) + \cdots + (10^n - 10^{n-1})}{10^n}\right] \\[10pt]
&= -\left(\frac{10^n - 1}{10^n}\right) \\[10pt]
&= -(1 - 10^{-n}) \\[10pt]
g(x) &= \frac{x^{r+1}}{r+1} - 10^n = \frac{x^{\displaystyle-(1 - 10^{-n})+1}}{-(1 - 10^{-n})+1} - 10^n \\[10pt]
\lim_{n\to\infty} g(x) &= \lim_{n\to\infty} \left(\frac{x^{\displaystyle-(1 - 10^{-n})+1}}{-(1 - 10^{-n})+1} - 10^n\right) \\[10pt]
&= \lim_{n\to\infty} \left(\frac{x^{10^{-n}}}{10^{-n}} - \frac{1}{10^{-n}}\right) \\[10pt]
&= \lim_{n\to\infty} 10^n \left(x^{\displaystyle\frac{1}{10^{n}}} - 1\right) \\[10pt]
\text{Let } m &= 10^n \Longrightarrow m\to\infty \text{ as } n\to\infty \\[10pt]
\lim_{n\to\infty} g(x) &= \lim_{m\to\infty} m(x^{\frac{1}{m}} - 1) \\[10pt]
&= \ln(x) \tag*{$\blacksquare$}
\end{align}$$
