lim_(x→1) ((4x lnx)/(x−1)) =??


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Question Number 97577 by Rio Michael last updated on 08/Jun/20

$\lim_{x \to 1} \frac{4 x \ln x}{x - 1} = ? ?$
Commented by Dwaipayan Shikari last updated on 23/Jun/20

$4 \lim_{x \to 1} \frac{x l o g (1 + x - 1)}{x - 1} = 4$
	Answered by smridha last updated on 08/Jun/20

	$\frac{0}{0} f o r m u s i n g L^{'} H \hat{o p i t a l} r u l e$ $4 \lim_{x \to 1} \frac{[1 + l n (x)]}{1} = 4$
	Commented by Rio Michael last updated on 08/Jun/20

	$thanks! please can you use another method$ $other thank L^{'} {Hopital}^{'} s rule$
	Commented by smridha last updated on 08/Jun/20

	$o k k$ $\lim_{x \to 1} \frac{4 x l n x}{x - 1} = \lim_{h \to 0} \frac{4 (h + 1) l n (h + 1)}{h}$ $= 4 \lim_{h \to 0} l n (h + 1) + 4 \lim_{h \to 0} \frac{l n (1 + h)}{h}$ $= 0 + 4 \times 1 = 4$
	Commented by Rio Michael last updated on 08/Jun/20

	$thank you$
	Answered by abdomathmax last updated on 08/Jun/20

	$changement x - 1 = t give \frac{4 xlnx}{x - 1} = \frac{4 (t + 1) \ln (1 + t)}{t}$ $t \to 1 \Rightarrow t \to 0 so \frac{4 (t + 1) \ln (1 + t)}{t} \sim \frac{4 (t + 1) t}{t}$ $= 4 (t + 1) \to 4 \Rightarrow \lim_{x \to 1} \frac{4 xlnx}{x - 1} = 4$
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