prove that lim_(x→0^+ ) ln x∙ln (1+x)=lim


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Question Number 34160 by candre last updated on 01/May/18

$p r o v e t h a t$ $\underset{x \to 0^{+}}{\lim l n} x \cdot \ln (1 + x) = \underset{x \to 1}{\lim l n} x \cdot \ln (1 + x)$
Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18

$s e e t h e g r a p h L H S l i m i t i s z e r o a l s o R H S l i m i t$ $a l s o z e r o$
Commented by abdo mathsup 649 cc last updated on 02/May/18

$w e h a v e {l i m}_{x \to 0^{+}} l n (x) l n (1 + x)$ $= {l i m}_{x \to 0^{+}} (x l n x) \frac{l n (1 + x)}{x} = 0 b e c a u s e {l i m}_{x \to 0^{+}} x l n (x) = 0$ ${l i m}_{x \to 0^{+}} \frac{l n (1 + x)}{x} = 1 f r o m a n o t h e r s i d e$ ${l i m}_{x \to 1} l n (x) . l n (1 + x) = l n (1) l n (2) = 0 b e c a u s e$ $t h e f u n c t i o n l n (x) i s c o n t i n u e a t 1 a n d 2 .$
	Answered by MJS last updated on 02/May/18

	$right side :$ $\ln 1 = 0$ $0 \ln 2 = 0$ $left side :$ $e^{\ln x \ln (x + 1)} = {(x + 1)}^{\ln x}$ $\lim_{x \to 0^{+}} {(x + 1)}^{\ln x} = 1^{- \infty} = 1 \Rightarrow$ $\Rightarrow \underset{x \to 0^{+}}{\lim l n} x \ln (x + 1) = 0$
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