Showing that the logarithm function is continuous in its domain boiled down to proving
$$\frac{x}{1+x}\le \ln(1+x)\le x \ \ \text{for all}\ x >-1.$$
There are quite a few proofs already online. Due to context, however, I need to justify this result
- without using continuity of logarithm or exponential functions;
- resorting to neither power series nor integration.
Is there a way?
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1$$
for $x>0$. Then, letting $x\to x+1$ we arrive at the coveted inequalities
$$\frac{x}{x+1}\le \log(1+x)\le x$$
And we are done!
Note that we can also arrive at the inequalities using the integral definition of the logarithm. Proceeding, we have
$$\log(x)=\int_1^x \frac{1}{t}\,dt$$
Then, it is easy to see that for $0<x$,
$$\frac{x-1}{x}=\int_1^x \frac1x\,dt\le \log(x)\le \int_1^x \frac{1}{1}\,dt=x-1$$
And continuity follows from the integral representation also since
$$\begin{align} |\log(x_2)-\log(x_1)|&\le \int_{x_1}^{x_2}\frac1{t}\,dt\\\\ &<\frac{x_2-x_1}{x_1}\\\\ &\to 0\,\,\text{as}\,\,x_1\to x_2 \end{align}$$