It's undefined. But $\lim_ {x \to 0^+} \ln x = -\infty$. If you extend the domain of $\ln$ to $ [0, \infty)$ and if you allow the extended function to take values in the "extended real numbers" $ [-\infty, \infty]$ or …

Mar 10, 2022 · My "proof" would be as follows: ε 2 = 0 ln 0 = ln ε 2 ln x 2 = 2 ln x ln 0 = 2 ln ε (a=b=ε) isn't really necessary. ln 0 = ln ab ln ab = ln a + ln b If ln (0) is always undefined, then it implies ln (ε) …

Dec 23, 2015 · Can we affirm that: $0 \\times \\ln(0) = \\ln(0^0) = \\ln(1) = 0$? The problem is $\\ln(0)$ is supposed to be undefined but it works

Feb 17, 2015 · When integrating $$\\int_{0}^{1}\\ln x\\,\\mathrm dx=\\ln(1)-1 - 0(\\ln(0)-0) = -1-0\\ln(0)$$ and so the integral should be indeterminate because $0\\ln(0)$ is ...

Note that $$\lim_ {x \to 0+} \ln x = -\lim_ {x \to 0+} \ln \frac {1} {x},$$ and for $0 < x < 1$ $$\ln \frac {1} {x} = \int_1^ {1/x}\frac {dt} {t} > \sum_ {k=1 ...

Apr 9, 2022 · The second answer is correct: $e^ {\ln 0}$ is undefined because $\ln 0$ is undefined. The definition of an inverse function says that $f\circ f^ {-1}=\text {id}$ on ...

Mar 25, 2021 · Why is $\ln 0\ne-\ln \infty$? They are equal: $\ln 0 = -\infty$ and $-\ln \infty = -\infty$. And thus, we can conclude that $\int_0^1\frac1xdx=\gamma+\int_1^\infty \frac1xdx$. Surprising, is not it, …

Sep 25, 2021 · I'm not clear on why you asked about $\frac {1} {\ln0}$, though, since that isn't what the original function is, and thus may not share discontinuities with it.

Nov 18, 2016 · The first step to this problem is to correct a phrasing. $\ln 0$ does not equal $-\infty$ and $\ln \infty$ does not equal $\infty$. 0 and infinity are not in the domain of ln. This is an important …

Aug 16, 2021 · The rule is, $0$ times any number is $0$. However, you still need the second thing to be a number. Because $\ln (0)$ is undefined, we can't multiply it by anything. You can think about it like …