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Question Number 134858 by mnjuly1970 last updated on 07/Mar/21

	Answered by mathmax by abdo last updated on 07/Mar/21

	$let f (x) = {lnxln}^{2} (1 - x) changement 1 - x = t give x = 1 - t (t \to 0^{+})$ $f (x) = \ln (1 - t) \ln^{2} (t) = g (t) we have$ $\ln^{^{'}} (1 - t) = \frac{- 1}{1 - t} = - (1 + t + o (t)) \Rightarrow \ln (1 - t) = - t - \frac{t^{2}}{2} + o (t^{2}) \Rightarrow$ $g (t) \sim (- t - \frac{t^{2}}{2}) \ln^{2} t \Rightarrow g (t) \sim - t \ln^{2} (t) - \frac{t^{2} \ln^{2} t}{2}$ $lnt = u \Rightarrow t = e^{u} \Rightarrow {tln}^{2} t = e^{u} u^{2} (u \to - \infty) \Rightarrow \lim u^{2} e^{u} = 0 \Rightarrow$ $also \lim tlnt = 0 \Rightarrow \lim_{t \to 0^{+}} g (t) = 0 = \lim_{x \to 1} f (x)$
	Answered by metamorfose last updated on 08/Mar/21

	$Q_{1} : \lim_{x \to 1^{-}} l n (x) {l n}^{2} (1 - x) = \lim_{x \to 1^{-}} \frac{l n (x)}{x - 1} (x - 1) {l n}^{2} (1 - x)$ $a n d, \lim_{x \to 1^{-}} \frac{l n (x)}{x - 1} = 1, \lim_{x \to 1^{-}} - (1 - x) {l n}^{2} (1 - x) = \lim_{t \to 0^{+}} - {t l n}^{2} (t) = 0$ $h e n c e, \lim_{x \to 1^{-}} l n (x) {l n}^{2} (1 - x) = 0 . . .$ $S a m e f o r Q_{2} : l e t y = 1 - x$ $t h e n \lim_{x \to 1^{-}} x = \lim_{x \to 0^{+}} y$ $s o : \lim_{y \to 0^{+}} {l n}^{2} (y) l n (1 - y) = \lim_{x \to 1^{-}} l n (x) {l n}^{2} (1 - x) = 0 . . .$
	Answered by Dwaipayan Shikari last updated on 08/Mar/21

	$\lim_{x \to 0} {l o g}^{2} (x) l o g (1 - x)$ $= 0$ $l o g (x) a p p r o a c h e s t o i n f i n i t y b u t a p p r o a c h i n g t o z e r o i s$ $m u c h f a s t e r (l o g (1 - x)) \to 0$
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