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Question Number 80670 by jagoll last updated on 05/Feb/20

lim_(x→π)  ((e^(sin x) −1)/(x−π))=?

$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{{e}^{\mathrm{sin}\:{x}} −\mathrm{1}}{{x}−\pi}=? \\ $$

Commented by jagoll last updated on 05/Feb/20

lim_(x→π)  ((cos x.e^(sin x) )/1)= −1

$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}.{e}^{\mathrm{sin}\:{x}} }{\mathrm{1}}=\:−\mathrm{1} \\ $$

Commented by abdomathmax last updated on 05/Feb/20

let f(x)=((e^(sinx) −1)/(x−π))  changement x−π=t give  f(x)=g(t)=((e^(sin(π+t)) −1)/t) =((e^(−sint) −1)/t)  ∼ ((e^(−t) −1)/t) ∼((1−t−1)/t)=−1 ⇒lim_(x→π)   f(x)=−1

$${let}\:{f}\left({x}\right)=\frac{{e}^{{sinx}} −\mathrm{1}}{{x}−\pi}\:\:{changement}\:{x}−\pi={t}\:{give} \\ $$$${f}\left({x}\right)={g}\left({t}\right)=\frac{{e}^{{sin}\left(\pi+{t}\right)} −\mathrm{1}}{{t}}\:=\frac{{e}^{−{sint}} −\mathrm{1}}{{t}} \\ $$$$\sim\:\frac{{e}^{−{t}} −\mathrm{1}}{{t}}\:\sim\frac{\mathrm{1}−{t}−\mathrm{1}}{{t}}=−\mathrm{1}\:\Rightarrow{lim}_{{x}\rightarrow\pi} \:\:{f}\left({x}\right)=−\mathrm{1} \\ $$

Answered by $@ty@m123 last updated on 05/Feb/20

Let x−π=t  x=π+t  As x→π, t=0  lim_(t→0)   ((e^(sin (π+t)) −1)/t)  lim_(t→0)   ((e^(−sin t) −1)/t)  lim_(t→0)   ((1−sin t+((sin^2 t)/(2!))+.....−1)/t)  lim_(t→0)   ((−sint)/t)  =−1

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

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