How can I prove that this limit is equal to the Natural Logarithm?

Solution 1:

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

Solution 2:

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 $$\begin{align} r &= -\sum_{k=0}^{n-1}\, 9 \cdot 10^{k-n} = -\frac{1}{10^n} \sum_{k=0}^{n-1}\, 9 \cdot 10^{k} \tag{$\lim_{n\to\infty} r = -1$} \\[10pt] &= -\left(\frac{9 + 90 + 900 + \cdots + 9\cdot 10^{n-1}}{10^n}\right) \\[10pt] &= -\left[\frac{(10-1) + (100-10) + (1000-100) + \cdots + (10^n - 10^{n-1})}{10^n}\right] \\[10pt] &= -\left(\frac{10^n - 1}{10^n}\right) \\[10pt] &= -(1 - 10^{-n}) \\[10pt] g(x) &= \frac{x^{r+1}}{r+1} - 10^n = \frac{x^{-(1 - 10^{-n})+1}}{-(1 - 10^{-n})+1} - 10^n \\[10pt] \lim_{n\to\infty} g(x) &= \lim_{n\to\infty} \left(\frac{x^{-(1 - 10^{-n})+1}}{-(1 - 10^{-n})+1} - 10^n\right) \\[10pt] &= \lim_{n\to\infty} \left(\frac{x^{10^{-n}}}{10^{-n}} - \frac{1}{10^{-n}}\right) \\[10pt] &= \lim_{n\to\infty} 10^n (x^{\frac{1}{10^{n}}} - 1) \\[10pt] \text{Let } m &= 10^n \Longrightarrow m\to\infty \text{ as } n\to\infty \\[10pt] \lim_{n\to\infty} g(x) &= \lim_{m\to\infty} m(x^{\frac{1}{m}} - 1) \\[10pt] &= \ln(x) \tag*{$\blacksquare$} \end{align}$$