calculate lim_(x→0) x ln(1−e^(sinx) )


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Question Number 41761 by math khazana by abdo last updated on 12/Aug/18

$c a l c u l a t e {l i m}_{x \to 0} x l n (1 - e^{s i n x})$
	Answered by alex041103 last updated on 12/Aug/18

	$A s x \to 0 t h e e x p r e s s i o n b e c o m e s o n e$ $o f t h e f o r m 0 \times \infty w h i c h i s e a s i l u$ $t r a n s f o r m e d i n \frac{\infty}{\infty} :$ $L = \lim_{x \to 0} \frac{l n (1 - e^{s i n x})}{1 / x} = L^{'} H o p i t a l$ $= \lim_{x \to 0} \frac{\frac{- e^{s i n x} c o s x}{1 - e^{s i n x}}}{- 1 / x^{2}} = \lim_{x \to 0} \frac{x^{2}}{1 - e^{s i n (x)}} {c o s x e}^{s i n x} =$ $= \lim_{x \to 0} \frac{x^{2}}{1 - e^{s i n x}} 1$ $N o w w e a p p l y L^{'} H o p i t a l u n t i l w e f i n d$ $a r e s u l t :$ $L = \lim_{x \to 0} \frac{2 x}{- e^{s i n (x)} c o s (x)} = \frac{2 \times 0}{- e^{s i n (0)} c o s (0)} =$ $= \frac{0}{1 \times 1} = 0$ $\Rightarrow {l i m}_{x \to 0} x l n (1 - e^{s i n x}) = 0$
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