$\epsilon$-$\delta$ proof that $\ln x$ is continuous everywhere on its domain

Question

How to use $\epsilon$-$\delta$ argument to show the global continuity of the function $x \mapsto \ln x$ on its domain?

For, if $c$ lies in the domain of $\ln x$ then $|\ln x - \ln c| = |\ln \frac{x}{c}|$. So how to relate the distance between the functional values to the distance between the arguments?

The distance between the functional values is $|\ln \frac xc|,$ as you just found. Now you just need to know how to ensure, for any given $\epsilon,$ that $|\ln \frac xc| < \epsilon.$ If you keep $x$ close enough to $c$, this will be true. And you don't need to find out the distance that is close enough; you just need to find a safe distance. — David K, Sep 02 '14 at 04:15

score 1 · Accepted Answer · answered Sep 02 '14 at 04:27

1

$x-c<\delta\Rightarrow x<c(\delta c^{-1}+1)\Rightarrow \ln x<\ln c + \ln(\delta c^{-1}+1)\Rightarrow \ln x-\ln c < \ln(\delta c^{-1}+1)$

Does this make some sense??

answered Sep 02 '14 at 04:27

So choose $\delta = c(e^{\epsilon}-1)$? – graydad Sep 02 '14 at 04:29
Yeah! it works. – Yes Sep 02 '14 at 04:30
Nice. $\epsilon -\delta$ moments like these are so satisfying. – graydad Sep 02 '14 at 04:31
@Comeseeconquer : You are sure?? – Sep 02 '14 at 04:35
@user166967 : yes.... :) :) – Sep 02 '14 at 04:36
My first attempt runs as follows: $x-c = \delta' < \delta$ so that $x = c + \delta'$ and hence $\ln \frac{x}{c} = \ln (1 + \frac{\delta'}{c}) < \ln (1 + \frac{\delta}{c}) \leq \epsilon$. – Yes Sep 02 '14 at 04:36
That sounds nice.... – Sep 02 '14 at 04:40
I just feel (even now) something not right about such argument but cannot tell, so I post the question here~ – Yes Sep 02 '14 at 04:42

$\epsilon$-$\delta$ proof that $\ln x$ is continuous everywhere on its domain

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