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$\displaystyle\lim_{x\to p} \ln(x) = \ln(p)$

Let $\epsilon>0$.

$|\ln(x)-\ln(p)| = \Big|\ln\Big(\frac{x}{p}\Big)\Big| $

Observe that:

$$\Big|\ln\Big(\frac{x}{p}\Big)\Big| < \epsilon \iff -\epsilon < \ln\Big(\frac{x}{p}\Big) < \epsilon \iff e^{-\epsilon} < \frac{x}{p} < e^\epsilon \iff pe^{-\epsilon} < x < pe^\epsilon \iff pe^{-\epsilon} - p < x - p < pe^\epsilon - p \iff p(e^{-\epsilon} - 1) < x - p < p(e^\epsilon-1)$$ $$ \iff -p(e^\epsilon - 1) < x - p < p(e^\epsilon - 1) \iff |x - p| < p(e^\epsilon - 1) $$

Note that $e^\epsilon - 1 \geq -e^{-\epsilon}+1$, since $e^x-1$ is convex, $-e^{-x}+1$ is concave and they share $y=x$ as tangent in $x=0$. Then $-(e^\epsilon - 1) \leq e^{-\epsilon}-1 $.

Then make $\delta = p(e^\epsilon - 1) > 0$.

So we have: $|x - p| < \delta \Rightarrow |\ln(x)-\ln(p)| < \epsilon$

I think my proof is ok, but I became confused after seeing what André Nicholas choosed for $\delta$ in this answer. Isn't this implying my choice for $\delta$ doesn't work?

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  • $\begingroup$ why not? isn't this true: $e^{-\epsilon}-1 \geq -(e^{\epsilon}-1)$? $\endgroup$ – creepyrodent Jun 7 '18 at 19:20
  • $\begingroup$ sorry, I still don't see the problem. I agree that $e^{-\epsilon} -1 < 0$, but $-(e^{\epsilon} - 1)$ is also negative... $\endgroup$ – creepyrodent Jun 7 '18 at 19:26
  • $\begingroup$ Sorry I must have brainfarted. Let me rephrase the problem. The '$\iff$' in this statement $$p(e^{-\epsilon} - 1) < x - p < p(e^\epsilon-1) \iff -p(e^\epsilon - 1) < x - p < p(e^\epsilon - 1)$$ is not true and should be an $\implies$, but you need the other direction. Indeed, we have $$-p(e^\epsilon - 1) \leqslant p(e^{-\epsilon} - 1) < p(e^\epsilon-1)$$ so that $p(e^{-\epsilon} - 1) < x-p \implies -p(e^\epsilon - 1) < x-p$, but not the other way round. $\endgroup$ – Fimpellizieri Jun 7 '18 at 19:34
  • $\begingroup$ and is this true? $p(e^{-\epsilon} - 1) < x-p < p(e^{\epsilon} - 1) \iff |x-p| < p(1-e^{-\epsilon})$? $\endgroup$ – creepyrodent Jun 7 '18 at 20:12
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    $\begingroup$ No. Think about what the LHS means and about what the RHS means, and see if they are the same. Also, notice that you do not need a sequence of iffs to show an $\epsilon-\delta$ limit. You need a sequence of implications from $|x-p|<\delta$ to $|f(x)-f(p)|<\epsilon$. $\endgroup$ – Fimpellizieri Jun 7 '18 at 20:18
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Note that, $$\exp\left|\ln(x) - \ln(p )\right| = \max\{\tfrac{x}{p},\tfrac{p}{x}\}$$

The only real issue we get with $\ln$ is at values near $0$, so to make sure we're away from it, we require $\delta < p/2$, then $|x-p|<\delta$ implies $p/2 < x$. By the triangle inequality \begin{align} \frac{x}{p} &\leq \left|\frac{p}{p}\right| + \left|\frac{x-p}{p}\right| < 1 + \frac{\delta}{p} < e^{\delta/p}\\ \frac{p}{x} &\leq \left|\frac{x}{x}\right| + \left|\frac{p-x}{x}\right| < 1 + \frac{\delta}{(p/2)} < e^{2\delta/p} \end{align}

Hence $$\exp\left|\ln(x) - \ln(p )\right| < \exp(2\delta/p)$$so if $\delta = \min\{p/2, \epsilon p /2\}$, then $|\ln(x)-\ln(p )| < \epsilon$.

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  • $\begingroup$ But isn't $|x/p| < e^{\epsilon}$ valid only when $x>p$? $\endgroup$ – creepyrodent Jun 7 '18 at 19:47
  • $\begingroup$ No, the triangle inequality holds in either case. If $|x-p| < \delta = p\epsilon$, then $$\left|\frac{x}{p}\right| < 1+\delta/p = 1 + \epsilon < 1+ \epsilon + \sum_{n=2}^\infty \frac{\epsilon^n}{n!} = e^{\epsilon}$$ $\endgroup$ – adfriedman Jun 7 '18 at 19:55
  • $\begingroup$ That part I understood, I mean this, at the begining: "$|\ln(x/p)| < \epsilon$ or equivalently $|x/p| < e^\epsilon $" $\endgroup$ – creepyrodent Jun 7 '18 at 20:00
  • $\begingroup$ Fixed it. Hopefully it is clearer now $\endgroup$ – adfriedman Jun 7 '18 at 21:03

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