Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 81096 by jagoll last updated on 09/Feb/20

lim_(x→3) (((sin x)/(sin 3)))^(1/(x−3)) =?

$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\mathrm{3}}\right)^{\frac{\mathrm{1}}{{x}−\mathrm{3}}} \:=? \\ $$

Commented by abdomathmax last updated on 09/Feb/20

let f(x)=(((sinx)/(sin3)))^(1/(x−3)) changement x−3=t give f(x)=g(t)=(((sin(t+3))/(sin3)))^(1/t) =(((sint cos3+cost sin3)/(sin3)))^(1/t) =(cotan3 sint +cost)^(1/t) =e^((1/t)ln(cost +cotan3 sint)) cost∼1−(t^2 /2) and sint ∼t ⇒ cost +cotan(3)sint ∼1−(t^2 /2) +cotan(3)t ⇒ ln(...)∼cotan(3)t−(t^2 /2) ⇒ (1/t)ln(cost +cotan(3))sint)∼cotan3 −(t/2)→cotan3 (t→0) ⇒lim_(t→0) g(t)=e^(cotan3) =e^((cos3)/(sin3)) =lim_(x→3) f(x)

$${let}\:{f}\left({x}\right)=\left(\frac{{sinx}}{{sin}\mathrm{3}}\right)^{\frac{\mathrm{1}}{{x}−\mathrm{3}}} \:{changement}\:{x}−\mathrm{3}={t}\:{give} \\ $$$${f}\left({x}\right)={g}\left({t}\right)=\left(\frac{{sin}\left({t}+\mathrm{3}\right)}{{sin}\mathrm{3}}\right)^{\frac{\mathrm{1}}{{t}}} =\left(\frac{{sint}\:{cos}\mathrm{3}+{cost}\:{sin}\mathrm{3}}{{sin}\mathrm{3}}\right)^{\frac{\mathrm{1}}{{t}}} \\ $$$$=\left({cotan}\mathrm{3}\:{sint}\:+{cost}\right)^{\frac{\mathrm{1}}{{t}}} \\ $$$$={e}^{\frac{\mathrm{1}}{{t}}{ln}\left({cost}\:+{cotan}\mathrm{3}\:{sint}\right)} \\ $$$${cost}\sim\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:{and}\:{sint}\:\sim{t}\:\Rightarrow \\ $$$${cost}\:+{cotan}\left(\mathrm{3}\right){sint}\:\sim\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:+{cotan}\left(\mathrm{3}\right){t}\:\Rightarrow \\ $$$${ln}\left(...\right)\sim{cotan}\left(\mathrm{3}\right){t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\left.\frac{\mathrm{1}}{{t}}{ln}\left({cost}\:+{cotan}\left(\mathrm{3}\right)\right){sint}\right)\sim{cotan}\mathrm{3}\:−\frac{{t}}{\mathrm{2}}\rightarrow{cotan}\mathrm{3} \\ $$$$\left({t}\rightarrow\mathrm{0}\right)\:\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} {g}\left({t}\right)={e}^{{cotan}\mathrm{3}} \:={e}^{\frac{{cos}\mathrm{3}}{{sin}\mathrm{3}}} ={lim}_{{x}\rightarrow\mathrm{3}} \:\:{f}\left({x}\right) \\ $$

Answered by john santu last updated on 09/Feb/20

limit form (1)^∞ we know that lim_(x→∞) (1+(1/x))^x = e let x−3 = t ⇒t→0 lim_(t→0) (((sin (t+3))/(sin 3)))^(1/t) =lim_(t→0) (1+(((sin (t+3)−sin 3)/(sin 3))))^(1/t) =e^(lim_(t→0) (((sin (t+3)−sin 3)/(t.sin 3)))) = e^(lim_(t→0) (((2sin ((t/2))cos ((t/2)+3))/(t.sin 3)))) = e^(cot 3) . the ans

$${limit}\:{form}\:\left(\mathrm{1}\right)^{\infty} \\ $$$${we}\:{know}\:{that}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \:=\:{e} \\ $$$${let}\:{x}−\mathrm{3}\:=\:{t}\:\Rightarrow{t}\rightarrow\mathrm{0} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\left({t}+\mathrm{3}\right)}{\mathrm{sin}\:\mathrm{3}}\right)^{\frac{\mathrm{1}}{{t}}} =\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\left(\frac{\mathrm{sin}\:\left({t}+\mathrm{3}\right)−\mathrm{sin}\:\mathrm{3}}{\mathrm{sin}\:\mathrm{3}}\right)\right)^{\frac{\mathrm{1}}{{t}}} \\ $$$$={e}^{\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\left({t}+\mathrm{3}\right)−\mathrm{sin}\:\mathrm{3}}{{t}.\mathrm{sin}\:\mathrm{3}}\right)} = \\ $$$${e}^{\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{2sin}\:\left(\frac{{t}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{t}}{\mathrm{2}}+\mathrm{3}\right)}{{t}.\mathrm{sin}\:\mathrm{3}}\right)} =\:{e}^{\mathrm{cot}\:\mathrm{3}} \:.\:{the}\:{ans} \\ $$

Commented by jagoll last updated on 09/Feb/20

waw thank you sir

$${waw}\:{thank}\:{you}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com