Question Number 190300 by mnjuly1970 last updated on 31/Mar/23
$$ \\ $$$$\:\:\:\mathrm{lim}_{\:{x}\rightarrow\mathrm{0}} \frac{\:{sin}\left({x}\right)\:−{tan}\left({x}\right)}{{x}^{\:\mathrm{3}} } \\ $$$$\:\:\:\:\:{a}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:{b}:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:{c}:\:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:{d}:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$
Answered by som(math1967) last updated on 31/Mar/23
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{sinx}\left({cosx}−\mathrm{1}\right)}{{x}.{x}^{\mathrm{2}} {cosx}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sinx}}{{x}}×\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{2}} {cosx}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{2}×\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{{cosx}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\left({c}\right) \\ $$
Commented by mnjuly1970 last updated on 31/Mar/23
$${very}\:{nice}\:{sir}\:.. \\ $$
Answered by cortano12 last updated on 31/Mar/23
$$\:\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{x}\right)−\mathrm{tan}\:\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{3}} }\: \\ $$$$\:\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)}{\mathrm{x}^{\mathrm{3}} } \\ $$$$\:\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{x}+\mathrm{1}\right)}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 31/Mar/23
$$\:{thx}\:{alot}\:\:{sir}\:\:{cortano} \\ $$