Question Number 111103 by bemath last updated on 02/Sep/20
![(√(bemath)) lim_(x→0) ((ln (sin 3x))/(ln (sin 8x))) ? [ Without L′Hopital ]](https://www.tinkutara.com/question/Q111103.png)
$$\:\:\sqrt{\mathrm{bemath}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{3x}\right)}{\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{8x}\right)}\:? \\ $$$$\left[\:\mathrm{Without}\:\mathrm{L}'\mathrm{Hopital}\:\right] \\ $$$$ \\ $$
Answered by Lordose last updated on 02/Sep/20
$$\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{about}}\:\boldsymbol{\mathrm{series}}…\: \\ $$
Commented by bemath last updated on 02/Sep/20
$$\mathrm{yes}.. \\ $$
Answered by bobhans last updated on 02/Sep/20
$$\mathrm{If}\:\mathrm{by}\:\mathrm{L}'\mathrm{Hopital}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3cot}\:\mathrm{3x}}{\mathrm{8cot}\:\mathrm{8x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\:\mathrm{tan}\:\mathrm{8x}}{\mathrm{8}\:\mathrm{tan}\:\mathrm{3x}}\:=\:\mathrm{1} \\ $$
Answered by Dwaipayan Shikari last updated on 02/Sep/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{log}\left({sin}\mathrm{3}{x}\right)}{{log}\left({sin}\mathrm{8}{x}\right)}=\frac{{log}\left(\mathrm{3}{x}\right)}{\mathrm{3}{x}}.\frac{\mathrm{8}{x}}{{log}\left(\mathrm{8}{x}\right)}=\frac{\mathrm{3}{x}−\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}}{\mathrm{3}{x}}.\frac{\mathrm{8}{x}}{\mathrm{8}{x}−\frac{\mathrm{64}{x}^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{1} \\ $$$$\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}\rightarrow\mathrm{0}\:,\mathrm{32}{x}^{\mathrm{2}} \rightarrow\mathrm{0} \\ $$$${sinx}\rightarrow{x} \\ $$