Question Number 106388 by bemath last updated on 05/Aug/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{4x}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{4}} } \\ $$
Answered by bemath last updated on 05/Aug/20
Answered by Dwaipayan Shikari last updated on 05/Aug/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}^{\mathrm{2}} {x}−{cos}\mathrm{4}{xsin}^{\mathrm{2}} {x}}{{x}^{\mathrm{4}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}^{\mathrm{2}} {x}\left(\mathrm{1}−{cos}\mathrm{4}{x}\right)}{{x}^{\mathrm{4}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{sin}^{\mathrm{2}} {x}.{sin}^{\mathrm{2}} \mathrm{2}{x}}{{x}^{\mathrm{4}} }=\frac{\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{2}{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{4}} }=\mathrm{8}\:\:\:\:\:\:\left({sinx}\rightarrow{x}\right) \\ $$