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Question Number 143339 by bramlexs22 last updated on 13/Jun/21

Commented by justtry last updated on 13/Jun/21

$$maybe0$$

Answered by mathmax by abdo last updated on 13/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{10\mathrm{x}}^2\mathrm{sinx}-{2\mathrm{xsin}}^2\left(3\mathrm{x}\right)+{\sin }^3\left(2\mathrm{x}\right)}{{\sin }^2\left(2\mathrm{x}\right)}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{10\mathrm{x}}^2\mathrm{sinx}-\mathrm{x}(1-\cos \left(6\mathrm{x}\right))+\sin \left(2\mathrm{x}\right)\frac{1-\cos \left(4\mathrm{x}\right)}{2}}{\frac{1-\cos \left(4\mathrm{x}\right)}{2}}$$$$=\frac{{20\mathrm{x}}^2\mathrm{sinx}-2\mathrm{x}+2\mathrm{xcos}\left(6\mathrm{x}\right)+\sin \left(2\mathrm{x}\right)-\sin \left(2\mathrm{x}\right)\cos \left(4\mathrm{x}\right)}{1-\cos \left(4\mathrm{x}\right)}$$$$\mathrm{we}\mathrm{have}\mathrm{sinx}\sim \mathrm{x}-\frac{{\mathrm{x}}^3}{6}\Rightarrow \sin \left(2\mathrm{x}\right)\sim 2\mathrm{x}-\frac{{8\mathrm{x}}^3}{6}=2\mathrm{x}-\frac{4}{3}{\mathrm{x}}^3$$$$\mathrm{cosu}=1-\frac{{\mathrm{u}}^2}{2}+\frac{{\mathrm{u}}^4}{4!}\Rightarrow \cos \left(6\mathrm{x}\right)\sim 1-{18\mathrm{x}}^2+\frac{6^4{\mathrm{x}}^4}{4!}$$$$\cos \left(4\mathrm{x}\right)\sim 1-{8\mathrm{x}}^2+\frac{4^4{\mathrm{x}}^4}{4!}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{{20\mathrm{x}}^2(\mathrm{x}-\frac{{\mathrm{x}}^3}{6})-2\mathrm{x}+2\mathrm{x}(1-{18\mathrm{x}}^2+\frac{6^4{\mathrm{x}}^4}{4!})+2\mathrm{x}-\frac{4}{3}{\mathrm{x}}^3-(2\mathrm{x}-\frac{4}{3}{\mathrm{x}}^3)(1-{8\mathrm{x}}^2+\frac{4^4{\mathrm{x}}^4}{4!})}{{8\mathrm{x}}^2-\frac{4^4{\mathrm{x}}^4}{4!}}$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{{20\mathrm{x}}^3-\frac{10}{3}{\mathrm{x}}^5+2\mathrm{x}-{36\mathrm{x}}^3+2.\frac{6^4}{4!}{\mathrm{x}}^5-\frac{4}{3}{\mathrm{x}}^3-(2\mathrm{x}-{16\mathrm{x}}^3+2\frac{4^4}{4!}{\mathrm{x}}^5-\frac{4}{3}{\mathrm{x}}^3+\frac{32}{3}{\mathrm{x}}^5-\frac{4^5}{3.4!}{\mathrm{x}}^7)}{{8\mathrm{x}}^2-\frac{4^4}{4!}{\mathrm{x}}^4}$$$$\mathrm{rdst}\mathrm{to}\mathrm{finish}\mathrm{the}\mathrm{calculus}....$$$$$$

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