lim_(x→0^+ ) xln(e^x −1)


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Question Number 205916 by mathzup last updated on 02/Apr/24

${l i m}_{x \to 0^{+}} x l n (e^{x} - 1)$
	Answered by mathzup last updated on 03/Apr/24
	$lim_(x→0^+ ) xln(e^x −1) =lim_(x→0^+ ) x{ln(e^x −1)−lnx +lnx} =lim_(x→0^+ ) xln(((e^x −1)/x))+xlnx lim_(x→0^+ ) xln(((e^x −1)/x))=0.ln(1)=0 lim_(x→0^+ ) xlnx =0 ⇒ lim_(x→0) xln(e^x −1)=0$
	${l i m}_{x \to 0^{+}} x l n (e^{x} - 1)$ $= {l i m}_{x \to 0^{+}} x {l n (e^{x} - 1) - l n x + l n x}$ $= {l i m}_{x \to 0^{+}} x l n (\frac{e^{x} - 1}{x}) + x l n x$ ${l i m}_{x \to 0^{+}} x l n (\frac{e^{x} - 1}{x}) = 0 . l n (1) = 0$ ${l i m}_{x \to 0^{+}} x l n x = 0 \Rightarrow$ ${l i m}_{x \to 0} x l n (e^{x} - 1) = 0$
	Commented by mathzup last updated on 03/Apr/24

	$a n o t h e r w a y w e d o t h e c h a n g e m e n t$ $e^{x} - 1 = t \Rightarrow e^{x} = 1 + t a n d x = l n (1 + t)$ $x \to 0 \Leftrightarrow t \to 0 a n d$ ${l i m}_{x \to 0^{+}} x l n (e^{x} - 1) = {l i m}_{t \to 0} l n (1 + t) l n t$ $= {l i m}_{t \to 0} \frac{l n (1 + t)}{t} \times {l i m}_{t \to 0} t l n t$ $= 1 \times 0 = 0$
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