Question Number 193248 by mnjuly1970 last updated on 08/Jun/23 | ||
$$ \\ $$$$\:\mathrm{L}=\:\mathrm{lim}_{\:{x}\rightarrow\mathrm{0}} \:\frac{\:\mathrm{sin}\left({x}\:\right)−\mathrm{arcsin}\left({x}\right)}{\mathrm{tan}\left({x}\right)−\:\mathrm{arctan}\left({x}\right)}=?\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\: \\ $$ | ||
Answered by MM42 last updated on 08/Jun/23 | ||
$${sinx}−{sin}^{−\mathrm{1}} {x}\:\sim\:−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}\:\:\:} \:\&\:\:\:{tanx}−{tan}^{−\mathrm{1}} {x}\:\sim\:\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \\ $$$$\Rightarrow{L}={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} }{\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} }\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ | ||