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I am confused, about how to proceed with this question, I got this far only and got stuck : For any $\varepsilon>0$, there exist $δ>0$ such that $$|\log(x) - \log(2)|<\varepsilon $$whenever $$\delta>|x-2|\implies \delta +2 > x > 2- \delta$$ I pick $\delta = 1$, then $$3<x<5$$ $$\log(3)<\log(x)<\log(5)$$ $$\log(x/2) < \log(5/2)$$

I think it may be wrong too, please help.

Afntu
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  • Welcome to MSE! As you are a new contributor to the site. I suggest you read some basic information about writing mathematics at this site, see, e.g. How can I format mathematics, There should be universal LaTeX/MathJax guide for sites supporting it, and MathJax basic tutorial and quick reference – Afntu Sep 01 '24 at 15:33
  • You don't want to set $\delta$ to anything spec. and you certainly don't want to set it as huge as $1$. If you are going from $\delta$ to $\epsilon$ the step after $\delta+2>x>2-\delta$ should be $\log(\delta+2)>\log x> \log(2-\delta)$ (assuming $\delta < 2$) It might be easy to go from $\epsilon$ to $\delta$. If $|log x-\log 2|< \epsilon$ then $\log 2-\epsilon < \log x< \log 2+\epsilon \iff 2\cdot e^{-\epsilon}=e^{\log 2-\epsilon}< x< e^{\log 2-\epsilon}=2\cdot e^\epsilon$. Cn you find some value of $\delta$ (in terms of $\epsilon$) that makes $2-\delta<2e^{-\epsilon}<x<2e^\epsilon<2+\delta$? – fleablood Sep 01 '24 at 15:50
  • This question is similar to: prove that lnx is continuous at 1 and at any positive real number a. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Afntu Sep 01 '24 at 15:54
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    You failed to tell us what was your definition of the $log$ function. It is commonly defined as 1/ the primitive of $x\mapsto\frac 1 x$ having value $0$ at $1$, or the inverse function of $x\mapsto e^x$. For the latter case you should give us what is your definition for the exponential function... – Serge Ballesta Sep 01 '24 at 16:06
  • Solving the inequalities $-\varepsilon<\log(x)-\log(2)<\varepsilon$ for $x$ gives you the $\delta$ in terms of $\varepsilon.$ After a bit of work inspired by your beginning work. – Steen82 Sep 01 '24 at 17:38

2 Answers2

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Given function is $f(x) = \ln(x)$, then $|\ln(x) - \ln(2)| = |\ln(x/2)|$, Now for any $\epsilon >0$,

$$|\ln(x/2)| < \epsilon \\ \iff e^{-\epsilon} < x/2 < e^{\epsilon} \\ \iff 2e^{-\epsilon} < x < 2e^{\epsilon} \\ \iff 2e^{-\epsilon} - 2 < x - 2 < 2e^{\epsilon} - 2$$

Let us choose $\delta = 2 - 2e^{-\epsilon} > 0$, also observe that $2 - 2e^{-\epsilon} \leq 2e^{\epsilon} - 2$. Hence for all $x$, satisfying $|x - 2| < \delta$ implies $|f(x) - f(2)| < \epsilon$.

Afntu
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"If I pick $\delta = 1$"

Why? Why pick $\delta = 1$. Why not just leave $\delta$ equaling $\delta$?

Then you get $2-\delta < x < 2+\delta$ and if we assume $2-\delta > 0$ (So we do restrict $0 < \delta < 2$) we get $\log(2-\delta) < \log x < \log(2+\delta)$.

Then your step of dividing by $2$ becomes $\log(\frac {2-\delta}2) < \log\frac x2 < \log(\frac {2+\delta}2)$ which is useful even without specific values.

and so

$\log(1-\frac \delta 2) < \log x - \log 2 < \log(1+\frac \delta 2)$.

Now we know $\log (1-\frac \delta 2) < 0$ and $\log(1+\frac \delta 2) > 0$ so if we can find a $\delta$ that would make

$-\epsilon \le \log(1-\frac \delta 2) < \log x-\log 2 < \log(1+\frac \delta 2)\le \epsilon$

we'd be done.

For $-\epsilon < \log(1-\frac \delta 2) < 0$ we need $e^{-\epsilon} < 1-\frac \delta 2$ or in other words: $\delta < 2(1-e^{-\epsilon})$.

And for $\log(1+\frac \delta 2) < \epsilon$ we need $1+\frac \delta 2 < e^{\epsilon}$ or $\delta < 2(e^\epsilon -1)$.

So as long as we pick $\delta$ so that $0 < \delta \le \min(2(1-e^{-\epsilon}), 2(e^\epsilon -1))$, then picking $x$ so that $|x-2| < \delta\le \min(2(1-e^{-\epsilon}), 2(e^\epsilon -1))$ we will have:

$-\epsilon \le \log(1-\frac \delta 2) < \log x-\log 2 < \log(1+\frac \delta 2)< \epsilon$ and so

$|\log x-\log 2| < \epsilon$

fleablood
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