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can anyone tell me, if this proof is correct:

for every $\epsilon>0$ there is a $\delta>0$ such that whenever $|x-1| < \delta$, then $|f(x)-f(1)| < \epsilon$. $|\ln(x)-\ln(1)|=|\ln(x)| < |\ln(0.5)| < \epsilon$ So: $δ=0.5$, $\epsilon > |\ln(0.5)|$

Keen-ameteur
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Kluse
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  • No, that is not correct. Example, if $\epsilon= e^{-20}$, then $|1.25-1|<0.5$, but $|\ln 1.25| \approx 0.22 > 1\times 10^{-8} > e^{-20}$. The idea of an $\epsilon$-$\delta$ proof is that you are given a value for $\epsilon > 0$, and with that value fixed, you need to come up with a value for $\delta$ related to $\epsilon$.

    You want $|\ln x| < \epsilon$, so $\ln x < \epsilon$ when $x \ge 1$ and $\ln x > -\epsilon$ when $x < 1$. So, if $x \ge 1$, then you need $x < e^\epsilon$ and if $x < 1$, you need $x > e^{-\epsilon}$. So, $\delta < \text{min}\left(e^\epsilon-1, 1-e^{-\epsilon}\right)$.

     – SlipEternal Jun 19 '18 at 16:39
  • Probably one of a million: https://math.stackexchange.com/questions/2811717/is-this-epsilon-delta-proof-of-lim-x-to-p-lnx-lnp-correct/2811743?noredirect=1#comment5797703_2811743 – adfriedman Jun 19 '18 at 17:04

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