Question Number 59542 by Mikael_Marshall last updated on 11/May/19
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\mathrm{ln}\left(\mathrm{cosx}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$${I}\:{have}\:{a}\:{doubt} \\ $$
Commented by kaivan.ahmadi last updated on 11/May/19
$${hop} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}\:} \:\frac{−{sinx}}{\mathrm{2}{xcosx}}\:\:={lim}_{{x}\rightarrow\mathrm{0}\:\:} \frac{−{tgx}}{\mathrm{2}{x}}= \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{−\left(\mathrm{1}+{tg}^{\mathrm{2}} {x}\right)}{\mathrm{2}}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Mikael_Marshall last updated on 11/May/19
$${oh}\:{thank}\:{you}\:{Sir} \\ $$
Commented by maxmathsup by imad last updated on 11/May/19
$${cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\:\:{and}\:\:{ln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\sim−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\frac{{ln}\left({cosx}\right)}{{x}^{\mathrm{2}} }\:\sim−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{ln}\left({cosx}\right)}{{x}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Mikael_Marshall last updated on 11/May/19
$${thank}\:{you}\:{Sir} \\ $$