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Question-195093




Question Number 195093 by Braulio last updated on 24/Jul/23
Answered by cortano12 last updated on 24/Jul/23
Answered by MM42 last updated on 24/Jul/23
hop→lim_(x→π)  (((1/2)cos(x/2)−sinx)/(2sinxcosx−sinx))  =lim_(x→π)  ((cos(x/2)((1/2)−2sin(x/2)))/(cos(x/2)(4sin(x/2)cosx−2sin(x/2))))   =(1/4) ✓
$${hop}\rightarrow{lim}_{{x}\rightarrow\pi} \:\frac{\frac{\mathrm{1}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}−{sinx}}{\mathrm{2}{sinxcosx}−{sinx}} \\ $$$$={lim}_{{x}\rightarrow\pi} \:\frac{{cos}\frac{{x}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}\right)}{{cos}\frac{{x}}{\mathrm{2}}\left(\mathrm{4}{sin}\frac{{x}}{\mathrm{2}}{cosx}−\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}\right)}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\checkmark \\ $$

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