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I have a question concerning the proof of the continuity of $f(x) = \ln x$.

I read in a comment by Pedro Tamaroff to ncmathsadist's answer to this question that this can be proved in two steps:

  1. first by proving that $f(x)$ is continuous at $1$,
  2. and then by using $\ln (xy) = \ln(x) + \ln(y)$.

However, I do not really see how to do it.

How does it actually work?

Community
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Kolmin
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1 Answers1

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Assume you have shown that $f=\ln$ is continuous at $1$.

We will use the following relation, that holds for any $x_0 > 0$ $$ \ln(x_0+h) = \ln\left(x_0\left(1+\frac{h}{x_0}\right)\right) = \ln x_0 + \ln\left(1+\frac{h}{x_0}\right). \tag{1} $$

For an $(\varepsilon,\delta)$ proof: fix any $x_0>0$, and choose any $\varepsilon > 0$. Let $\delta_1$ be such that $$ \lvert\ln x - \ln 1\rvert \leq \varepsilon $$ whenever $\lvert x-1\rvert \leq \delta_1$ (this exists by continuity of $\ln$ at $1$). Now, set $\delta \stackrel{\rm def}{=} x_0\delta_1$.

For any $x>0$ such that $\lvert x-x_0\rvert \leq \delta$, we have by (1) $$\begin{align} \lvert \ln x - \ln x_0\rvert &= \lvert \ln (x_0+\underbrace{(x-x_0)}_{"h"}) - \ln x_0\rvert = \left\lvert \ln\left( 1+\frac{x-x_0}{x_0}\right)\right\rvert\\ &= \left\lvert \ln\left( 1+\frac{x-x_0}{x_0}\right) - \ln 1\right\rvert \leq \varepsilon \end{align}$$ the last inequality since $\left\lvert\frac{x-x_0}{x_0}\right\rvert \leq \frac{\delta}{x_0} = \frac{x_0\delta_1}{x_0} = \delta_1$.

So $\ln$ is continuous at $x_0$.

Clement C.
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  • is my answer wrong? – BCLC Jun 07 '16 at 02:27
  • Why would you think so? – Clement C. Jun 07 '16 at 02:28
  • Thanks a lot for the detailed answer. May I ask you why $\ln x_0 + \ln ( 1 + \frac{h}{x_0})$ implies continuity? – Kolmin Jun 07 '16 at 02:32
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    Read the rest: (1) is then used in the proof of continuity, to show the existence of a "good $\delta$" for arbitrary $x_0$ and $\varepsilon$. (I.e., (1) is not a full proof, it's the main building block in the proof that follows.) – Clement C. Jun 07 '16 at 02:33
  • I think I see. Hence (1) does not imply continuity straight away: it is rather something we use to establish continuity via $\epsilon -\delta$. But then, isn't it more straightforward to just prove the continuity of $f(x)$ by taking an arbitrary $x_0$ and then using $\epsilon -\delta$, without bothering with the initial extra step of establishing continuity at $1$? – Kolmin Jun 07 '16 at 02:37
  • [while you were editing, I was writing this comment] – Kolmin Jun 07 '16 at 02:38
  • It may be -- depending on how and if you manage to do so (the ease of which depends a lot on how the way $\ln$ was defined, which of its properties were established, and the set of tools/results you allow). – Clement C. Jun 07 '16 at 02:38
  • thus, you would say that this "double" step proof is actually more basic. – Kolmin Jun 07 '16 at 02:39
  • I will check the second part to see if it is everything clear to me. If not, maybe I will bother you again. – Kolmin Jun 07 '16 at 02:39
  • Again, depending on how $\ln$ was introduced, yes, it may be simpler to do it that way. (And regarding your latest comment: sure.) – Clement C. Jun 07 '16 at 02:40
  • ClementC. Idk. I was just guessing – BCLC Jun 07 '16 at 02:59
  • @ClementC. Indeed, I do have some questions: [a] Why do we need to have established the continuity at $1$ to obtain equation (1)? [b] I can see every step with one exception, namely how to connect $$\displaystyle \left\lvert \ln\left( 1+\frac{x-x_0}{x_0}\right) - \ln 1\right\rvert $$ and $$ \displaystyle \left\lvert\frac{x-x_0}{x_0}\right\rvert .$$ (mind that I see why the first expression is $\leq 0$ through the chain of implications you do from the second… I just don’t see how the two are related). – Kolmin Jun 07 '16 at 03:12
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    @Kolmin No need for continuity at $1$ to establish (1) (the phrasing of the first sentence is confusing, I may add a word to show better the structure). For your second question, write $u=1+\frac{x-x_0}{x_0}$, and then you have $\lvert \ln u - \ln 1\rvert$ and $\lvert u-1\rvert$. $\frac{x-x0}{x_0}=u-1$ is the thing that is bounded by $\delta_1$, for the continuity of $\ln$ at $1$. – Clement C. Jun 07 '16 at 03:15
  • Thanks a lot. Superuseful and clear! – Kolmin Jun 07 '16 at 03:21
  • You're welcome! – Clement C. Jun 07 '16 at 03:21