lim_(x→1^− ) lnx∙ln(1−x)=?


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Question Number 78122 by Tony Lin last updated on 14/Jan/20

$\lim_{x \to 1^{-}} l n x \cdot l n (1 - x) = ?$
Commented by msup trace by abdo last updated on 14/Jan/20

$c h a n g e m e n t 1 - x = t g i v e$ ${l i m}_{x \to 1^{-}} l n x l n (1 - x)$ $= {l i m}_{t \to 0^{+}} l n (1 - t) . l n (t)$ $= {l i m}_{t \to 0^{+}} (t l n t) \times \frac{l n (1 - t)}{t}$ $w e h a v e {l i m}_{t \to 0^{+}} (t l n t) = 0$ ${l i m}_{t \to 0^{+}} \frac{l n (1 - t)}{t} = - 1 \Rightarrow$ ${l i m}_{x \to 1^{-}} l n x l n (1 - x) = 0$
Commented by Tony Lin last updated on 14/Jan/20

$t h a n k s s i r$
Commented by msup trace by abdo last updated on 14/Jan/20

$y o u a r e w e l c o m e$
	Answered by john santu last updated on 14/Jan/20

	$u s i n g b y s e r i e s ?$
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