Question-32376

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 \to 0} \frac{e^{x} - e^{\sin x}}{x - \sin x}

\lim_{x \to 0} \frac{(1 + x + \frac{x^{2}}{2!} + \dots) - (1 + \sin x + \frac{\sin^{2} x}{2!} + \dots)}{x - \sin x}

\lim_{x \to 0} \frac{(x - \sin x) + \frac{1}{2!} (x^{2} - \sin^{2} x) + \dots .}{x - \sin x}

\lim_{x \to 0} \frac{(x - \sin x) {1 + \frac{1}{2!} (x + \sin x) + \dots .}}{x - \sin x}

\lim_{x \to 0} {1 + \frac{1}{2!} (x + \sin x) + \dots .}

= 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