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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?

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    $\begingroup$ Why was this question downvoted? $\endgroup$ – BCLC Jun 7 '16 at 2:24
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    $\begingroup$ Just wondering, why is this question so downvoted? $\endgroup$ – Kolmin Jun 7 '16 at 2:47
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    $\begingroup$ @Kolmin well, the title is not totally clear, but that's not enough of a reason really. $\endgroup$ – YoTengoUnLCD Jun 7 '16 at 4:27
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    $\begingroup$ You should provide some more details perhaps about how you define the symbol $\log x$. The answer is crucially dependent on this information. $\endgroup$ – Paramanand Singh Jun 7 '16 at 8:13
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    $\begingroup$ Closely related question: Functional equation $f(xy)=f(x)+f(y)$ and continuity $\endgroup$ – Martin Sleziak Jun 7 '16 at 16:46
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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 $x_0$.

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  • $\begingroup$ is my answer wrong? $\endgroup$ – BCLC Jun 7 '16 at 2:27
  • $\begingroup$ Why would you think so? $\endgroup$ – Clement C. Jun 7 '16 at 2:28
  • $\begingroup$ Thanks a lot for the detailed answer. May I ask you why $\ln x_0 + \ln ( 1 + \frac{h}{x_0})$ implies continuity? $\endgroup$ – Kolmin Jun 7 '16 at 2:32
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    $\begingroup$ 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.) $\endgroup$ – Clement C. Jun 7 '16 at 2:33
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    $\begingroup$ @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 $1$. $\endgroup$ – Clement C. Jun 7 '16 at 3:15

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