Find $\lim_\limits{x\to 1^-}{\ln x\cdot \ln(1-x)}$. I can't even start because I don't really know what $x\to 1^-$ means. If you know what it means it would really help me. I would as well if you helped me with finding the limit, hinted or thorough.
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I did realize that tending to $1$ is problematic because of $ln(1-x)$, but is it possible to define a limit that way? – Meitar Abarbanel Jan 29 '15 at 00:02
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3$x \to 1 - 0$ means $x$ is approaching $1$ from left. Like $0.9, 0.99, 0.999 \dots$. This ensures that $\ln{(1- x)}$ is well defined. – Jan 29 '15 at 00:02
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2In such a situation, the rule of L'Hospital may help... – russoo Jan 29 '15 at 00:04
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I shall try. But, first, I have to make it a quotient. – Meitar Abarbanel Jan 29 '15 at 00:10
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$$\frac{\log(1-x)}{\frac1{\log x}}\stackrel{l'H}\rightarrow\frac{-\frac1{1-x}}{-\frac{\frac1x}{\log^2x}}=\frac{x\log^2 x}{1-x}\stackrel{l'H}\rightarrow\frac{\log^2x+2\log x}{-1}\xrightarrow[x\to1^-]{}0 $$
Timbuc
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Thanks. Turned out I accidentally differentiated ${1\over \ln x}$ incorrectly . – Meitar Abarbanel Jan 29 '15 at 00:18
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With equivalents, it's very short: set $x=1-u\enspace(u>0)$. Then $\ln x\ln(1-x)=\ln(1-u)\ln u$.
As $\ln(1-u)\sim_0 -u$, we have $\,\,\ln(1-u)\ln u\sim_0-u\ln u$, which tends to $0$ as $u$ tends to $0$.
Hence $$\lim_{x\to 1_-}\ln x\,\ln(1-x)=\lim_{u\to 0_+}\ln(1-u)\ln u=0.$$
Bernard
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is $\lim_{u \to 0+} \ln(1-u) \ln u$ not an indeterminate form $0.\infty?$ – abel Jan 29 '15 at 00:20
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$f~ag $ roughly means that $\lim{x\to a}\frac fg=1$. Ver often, an equivalent is given by the first non-zero term of a Taylor expansion. Here, for instance, $\ln(1-u)=-u-\dfrac{u^2}2+\dots$, hence $\ln(1-u)\sim_0 u$. It's a powerful tool to replace complicated expressions by their meaningful (simpler) part. – Bernard Jan 29 '15 at 00:29
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ok i get it but i think it will be helpful to include an intermediate step $\ln(1- u) \ln u = u \ln u + \cdots$ – abel Jan 29 '15 at 00:34
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I don't think so: it's a well known rule that equivalence is compatible with multiplication and division (but not with addition). You can take a look at this Wikipedia notice (sorry it's in French, but the English version gives almost no details). – Bernard Jan 29 '15 at 00:47
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$$ \lim_{x\uparrow1} \ln x \ln(1-x) = \lim_{x\uparrow1} \frac{\ln x}{x-1}\cdot\frac{\ln(1-x)}{1/(1-x)}. $$
Apply L'Hopital's rule to both fractions and you've got it. (In the first one, the numerator and denominator both approach $0$; in the second, each approaches either $+\infty$ or $-\infty$.)
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here is a way to avoid l'hospitals but will use the fact $\lim_{n \to \infty} \dfrac{\ln n}{n} = 0$ let us do a change of variable $x = 1 - \dfrac{1}{n}.$ so $$\ln x \ln (1- x) = \ln(1 - 1/n) \ln(1/n)=\ln n(-\ln(1- 1/n)=\ln n\int_{1-1/n}^1\dfrac{dt}{t}$$
we can use the bound $\dfrac{1}{n} \le \int_{1-1/n}^1\dfrac{dt}{t} \le \dfrac{1}{n-1}$ to conclude $$ \dfrac{\ln n}{n} \le \ln x \ln (1- x) \le \dfrac{\ln n}{n-1} $$
therefore the $$\lim_{x \to 1_-} \ln x \ln (1- x) = 0$$
abel
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To prove that \begin{equation*} \lim_{x\rightarrow 1^{-}}\ln x\ln (1-x)=0, \end{equation*} it is possible to avoid, L'Hospital's rule, Taylor series and also to avoid the limits \begin{equation*} \lim_{u\rightarrow 0}u\ln u=0,\ \ \ \ \text{and}\ \ \ \ \ \ \ \ \lim_{n\rightarrow +\infty }\frac{\ln n}{n}=0. \end{equation*}
It suffices to use the well-known inequality \begin{equation*} 0\leq \ln x\ln (1-x)\leq \sqrt{x(1-x)},\ \ x\in (0,1). \end{equation*} and the squeeze theorem.
PS. An elegant proof of the well known inequality cited above maybe found here:
Idris Addou
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