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

we have e^x ∼1+x +(x^2 /2) for x∈V(0) alsowehave e^(sinx) =e^(x−(x^3 /(3!)) +0(x^5 )) ∼1+x −(x^3 /(3!)) ⇒ e^x −e^(sinx) ∼(x^3 /(3!)) sinx ∼ x−(x^3 /(3!)) ⇒ x−sinx ∼ (x^3 /(3!)) ⇒ ((e^x −e^(sinx) )/(x−sinx)) ∼ 1 ⇒ lim_(x→o) ((e^x −e^(sinx) )/(x −sinx)) =1 .

$${we}\:{have}\:{e}^{{x}} \:\sim\mathrm{1}+{x}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{for}\:{x}\in{V}\left(\mathrm{0}\right)\:{alsowehave} \\ $$$${e}^{{sinx}} \:={e}^{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+\mathrm{0}\left({x}^{\mathrm{5}} \right)} \:\sim\mathrm{1}+{x}\:−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:\Rightarrow\:{e}^{{x}} \:−{e}^{{sinx}} \:\sim\frac{{x}^{\mathrm{3}} }{\mathrm{3}!} \\ $$$${sinx}\:\sim\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:\:\Rightarrow\:{x}−{sinx}\:\sim\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:\Rightarrow \\ $$$$\:\frac{{e}^{{x}} \:−{e}^{{sinx}} }{{x}−{sinx}}\:\sim\:\:\mathrm{1}\:\:\Rightarrow\:{lim}_{{x}\rightarrow{o}} \:\:\frac{{e}^{{x}} \:−{e}^{{sinx}} }{{x}\:−{sinx}}\:=\mathrm{1}\:\:. \\ $$

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

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{{e}^{{x}} −{e}^{\mathrm{sin}\:{x}} }{{x}−\mathrm{sin}\:{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+…\right)\:−\left(\mathrm{1}+\mathrm{sin}\:{x}+\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{2}!}+…\right)}{{x}−\mathrm{sin}\:{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\left({x}−\mathrm{sin}\:{x}\right)+\frac{\mathrm{1}}{\mathrm{2}!}\left({x}^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} {x}\right)+….\:}{{x}−\mathrm{sin}\:{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\left({x}−\mathrm{sin}\:{x}\right)\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!}\left({x}+\mathrm{sin}{x}\right)+….\right\}\:}{{x}−\mathrm{sin}\:{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!}\left({x}+\mathrm{sin}{x}\right)+….\right\} \\ $$$$=\mathrm{1} \\ $$

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