Question Number 174683 by infinityaction last updated on 08/Aug/22
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{{x}^{\mathrm{2}} }{\mathrm{sin}{x}\:\mathrm{tan}{x}\:}\right] \\ $$$$\left[\centerdot\right]\:{greatest}\:{integer}\:{function} \\ $$
Answered by a.lgnaoui last updated on 09/Aug/22
$$ \\ $$$$\left[\frac{\mathrm{x}^{\mathrm{2}} \mathrm{cos}\:\mathrm{x}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\right]=\mathrm{cos}\:\mathrm{x}\left[\frac{\mathrm{1}}{\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} }}\right] \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{sin}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}}=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{cos}\:{x}×\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$
Commented by infinityaction last updated on 09/Aug/22
$${no} \\ $$
Answered by CElcedricjunior last updated on 09/Aug/22
$$\mathrm{1} \\ $$$$ \\ $$
Commented by infinityaction last updated on 09/Aug/22
$${no}\:{answer}\:{is}\:\mathrm{0} \\ $$