Question Number 42786 by maxmathsup by imad last updated on 02/Sep/18
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{1}−\frac{{x}}{{sinx}}}{{x}^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 30/Sep/18
$${we}\:{have}\:{sinx}\:=\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+{o}\left({x}^{\mathrm{3}} \right)\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\frac{{sinx}\:−{x}}{{x}^{\mathrm{2}} {sinx}}\:\:\sim\:\:\frac{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}}{{x}^{\mathrm{3}} }\:\rightarrow−\frac{\mathrm{1}}{\mathrm{6}}\:\:{when}\:{x}\rightarrow\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−\frac{{x}}{{sinx}}}{{x}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{6}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sinx}−{x}}{{x}^{\mathrm{2}} {sinx}}\:\:\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{cosx}−\mathrm{1}}{{x}^{\mathrm{2}} {cosx}+{sinx}.\mathrm{2}{x}}\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sinx}}{−{x}^{\mathrm{2}} {sinx}+{cosx}.\mathrm{2}{x}+{sinx}.\mathrm{2}+\mathrm{2}{xcosx}}\:\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{sinx}}{{x}}}{−{xsinx}+\mathrm{2}{cosx}+\mathrm{2}\frac{{sinx}}{{x}}+\mathrm{2}{cosx}} \\ $$$$=\frac{\mathrm{1}}{−\mathrm{0}+\mathrm{2}+\mathrm{2}+\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$