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Question Number 32376 by rsen3579@gmail.com last updated on 24/Mar/18

Commented by abdo imad last updated on 24/Mar/18

$w e h a v e e^{x} \sim 1 + x + \frac{x^{2}}{2} f o r x \in V (0) a l s o w e h a v e$ $e^{s i n x} = e^{x - \frac{x^{3}}{3!} + 0 (x^{5})} \sim 1 + x - \frac{x^{3}}{3!} \Rightarrow e^{x} - e^{s i n x} \sim \frac{x^{3}}{3!}$ $s i n x \sim x - \frac{x^{3}}{3!} \Rightarrow x - s i n x \sim \frac{x^{3}}{3!} \Rightarrow$ $\frac{e^{x} - e^{s i n x}}{x - s i n x} \sim 1 \Rightarrow {l i m}_{x \to o} \frac{e^{x} - e^{s i n x}}{x - s i n x} = 1 .$
	Answered by $@ty@m last updated on 24/Mar/18
	$lim_(x→0) ((e^x −e^(sin x) )/(x−sin x)) lim_(x→0) (((1+x+(x^2 /(2!))+...) −(1+sin x+((sin^2 x)/(2!))+...))/(x−sin x)) lim_(x→0) (((x−sin x)+(1/(2!))(x^2 −sin^2 x)+.... )/(x−sin x)) lim_(x→0) (((x−sin x){1+(1/(2!))(x+sinx)+....} )/(x−sin x)) lim_(x→0) {1+(1/(2!))(x+sinx)+....} =1$
	$\lim_{x \to 0} \frac{e^{x} - e^{\sin x}}{x - \sin x}$ $\lim_{x \to 0} \frac{(1 + x + \frac{x^{2}}{2!} + . . .) - (1 + \sin x + \frac{\sin^{2} x}{2!} + . . .)}{x - \sin x}$ $\lim_{x \to 0} \frac{(x - \sin x) + \frac{1}{2!} (x^{2} - \sin^{2} x) + . . . .}{x - \sin x}$ $\lim_{x \to 0} \frac{(x - \sin x) {1 + \frac{1}{2!} (x + \sin x) + . . . .}}{x - \sin x}$ $\lim_{x \to 0} {1 + \frac{1}{2!} (x + \sin x) + . . . .}$ $= 1$
	Answered by saru53424@gmail.com last updated on 24/Mar/18

	Answered by saru53424@gmail.com last updated on 24/Mar/18

	Answered by saru53424@gmail.com last updated on 24/Mar/18

	$x + 1 = 0$
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