Question Number 192433 by cortano12 last updated on 18/May/23
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{ln}\:\left(\mathrm{3x}+\mathrm{1}\right)\right)}{\mathrm{3x}^{\mathrm{6}} −\mathrm{tan}\:\left(\mathrm{3x}^{\mathrm{2}} \right)}=? \\ $$
Answered by mehdee42 last updated on 18/May/23
$${tip}\::{if}\:\:{u}\rightarrow\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−{cosu}\sim\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} \:\:\&\:\:{ln}\left({u}+\mathrm{1}\right)\sim{u}\:\:\:\:\&\:\:{tanu}\sim{u} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−{co}\left({ln}\left(\mathrm{3}{x}+\mathrm{1}\right)\right)}{\mathrm{3}{x}^{\mathrm{6}} −{tan}\left(\mathrm{3}{x}^{\mathrm{2}} \right)}={lim}_{{x}\rightarrow\mathrm{0}} \frac{\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{3}{x}+\mathrm{1}\right)\right)^{\mathrm{2}} }{\mathrm{3}{x}^{\mathrm{6}} −\mathrm{3}{x}^{\mathrm{2}} }\:={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}}{\mathrm{3}{x}^{\mathrm{2}} \left({x}^{\mathrm{4}} −\mathrm{1}\right)}=−\frac{\mathrm{3}}{\mathrm{2}}\:\checkmark \\ $$