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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?

Martin Sleziak
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Linear Christmas
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  • What tools can one use here? – Mark Viola Jul 17 '16 at 16:58
  • Are you allowed to use differentiation? Ah wait no if not allowed to use continous we can't really use differentiability which is stronger... right? – mathreadler Jul 17 '16 at 17:00
  • Technically not. The problem is before the differentiation chapter of my textbook. Though I would be completely fine with a differentiation proof as long as it does not assume continuity. – Linear Christmas Jul 17 '16 at 17:02
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    What is your definition of the logarithm function? – Mark Viola Jul 17 '16 at 17:04
  • At least in my calculus treatment a long time ago differentiability requires continuity. Yet there are functions continuous everywhere but differentiable nowhere. Those functions are rather weird though. – mathreadler Jul 17 '16 at 17:06
  • If using the logarithm definition Dr.MV uses down below maybe it would be useful to rewrite left hand side as $1-\frac{1}{1+x}$ – mathreadler Jul 17 '16 at 17:08
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    See here Dr MV's own proof here which uses only the limit definition of the exponential function and Bernoulli's Inequality: http://math.stackexchange.com/questions/1589429/how-to-prove-that-logxx-when-x1/1590263#1590263 – smcc Jul 17 '16 at 17:09
  • @smcc Thank you for the reference. How did you find that so quickly? Most impressive. -Mark – Mark Viola Jul 17 '16 at 17:13
  • You linked to it a day or so ago in answering another question about the limit of $\ln x/x$ that I also answered. – smcc Jul 17 '16 at 17:15

1 Answers1

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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

$$\begin{align} |\log(x_2)-\log(x_1)|&\le \int_{x_1}^{x_2}\frac1{t}\,dt\\\\ &<\frac{x_2-x_1}{x_1}\\\\ &\to 0\,\,\text{as}\,\,x_1\to x_2 \end{align}$$

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Mark Viola
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  • The first half of your answer is exactly what I'm looking for. But if I were to in turn prove the Bernouilli's inequality for real numbers, wouldn't that require using derivatives anyway? – Linear Christmas Jul 17 '16 at 17:21
  • The proof of the version of Bernoulli's inequality with integer exponent does not require differentiation. See here: https://en.wikipedia.org/wiki/Bernoulli%27s_inequality#Proof_of_the_inequality – smcc Jul 17 '16 at 17:35
  • Bernoulli's Inequality as applied here, was for integer exponents. And its proof relies on elementary induction. – Mark Viola Jul 17 '16 at 17:35
  • Indeed, I was referring to the proof for real exponents. – Linear Christmas Jul 17 '16 at 17:38
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    If you can prove Bernoulli's Inequality for rational exponents, then by a density argument, you can extend this to the reals. But this is not relevant for the development herein. -Mark – Mark Viola Jul 17 '16 at 17:43
  • Thank you for sharing your insight! I can now move on to proving continuity for other main elementary functions. I have a feeling that the inequalities mentioned in this post will be most handy in the future to come. – Linear Christmas Jul 17 '16 at 17:46
  • You're welcome! My pleasure. And pleased to hear that this was useful!! -Mark – Mark Viola Jul 17 '16 at 17:48
  • @Dr.MV: Haven't you, by any chance, proven somewhere that $\ |\sin \frac{x}{2} | \leq |\sin x|$ with similar restrictions (i.e, no continuity)? ;) – Linear Christmas Jul 18 '16 at 14:05
  • To be fair, I now see that for my purposes it was enough to show $\ |\sin \frac{x}{2} | \leq |\sin x|$ is valid for some interval centered around $0$. But instead the problem is circumvented altogether using a different inequality "proved" with a unit circle. I was initially reluctant to use such an argument due to this question. For my purposes, a geometry arguement is probably rigorous enough. – Linear Christmas Jul 18 '16 at 18:45
  • The inequality is obviously not true in general (e.g., For $x=\pi$, $|\sin(x)|=0$ and $|\sin(x/2)|=1$). – Mark Viola Jul 18 '16 at 19:35
  • Note that $f(x)=\sin(x)-\sin(x/2)=\sin(x/2)(2\cos(x/2)-1)$, which is greater than zero when $\cos(x/2)>1/2$ and $\sin(x/2)>0$ ( $0<x<2\pi/3$) or $\cos(x/2)<1/2$ and $\sin(x/2)<0$ ($0>x>-2\pi/3$) for $-\pi < x< \pi$. – Mark Viola Jul 18 '16 at 19:41