$\lim_{x\to1}\frac{\ln(x)}{x-1}$ and its strange graph

Question

I was studying exponential growth and noticed that

• $\ln(0.99) \approx −0.010050336$
• $\ln(0.999) \approx −0.0010005$
• $\ln(0.9999) \approx −0.0001 \ldots$, and also
• $\ln(0.9952) \approx −0.004811557$

It seems like $\ln(x) \approx x-1$ for $x\to1$. On a linear scale, this looks ordinary. My first question is if this theorem is true and how it is called. It seems only to hold for the natural logarithm, though.

To get more insight, I switched $x$ to a logarithmic scale, i.e.

$$f(x) = \frac{\ln(1-10^{-x})}{1-10^{-x}-1} = -\frac{\ln(1-10^{-x})}{10^{-x}}$$

This is how it looks.

The exact position depends on the base of the exponentation, which is 10 in this case.

Can someone explain this strange behaviour? Why is there a gap?

Answer 1

We can look at the Maclaurin series of log to see (for $x$ close to zero) that:$$\ln (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots$$

The interesting patterns you see are all artifacts of computation. Wolfram Alpha, or whatever software you use, represents numbers in a particular way, and that way leads to rounding errors which get increasingly magnified to produce the graphs you see.

We can calculate your function exactly: $$f(x)=-\frac{\ln (1-10^{-x})}{10^{-x}}=-\frac{-10^{-x}-(\frac{1}{2})10^{-2x}-\cdots}{10^{-x}}=1+(\frac{1}{2})10^{-x}+\cdots\approx 1+(\frac{1}{2})10^{-x}$$

Hence for $x>3$, $f(x)$ SHOULD be indistinguishable from $1$ in terms of a visual plot.

Answer 2

As: $$\ln(1+x)=x-x^2/2+x^3/3-x^4/4+\cdots\tag{$|x|<1$}$$ So: $$\ln(x)=(x-1)-(x-1)^2/2+(x-1)^3/3+\cdots\tag{$|x|\in(0,2)$}$$ So for $x-1\ll1\iff x\sim0$: $$\ln x\approx x-1$$ Also using L'Hospital: $$\lim_{x\to1}\frac{\ln x}{x-1}=\lim_{x\to1}\frac{1/x}{1-0}=1$$ Or substituting $x=1+y$: $$\lim_{x\to1}\frac{\ln x}{x-1}=\lim_{y\to0}\frac{\ln (1+y)}{y}=1$$ by using the predetermined very common identity of limit of $\ln(1+y)$ over $y$ around zero.