Question Number 185510 by mathlove last updated on 23/Jan/23
Answered by mahdipoor last updated on 23/Jan/23
$$\mathrm{6}{x}−\mathrm{8}{sin}\left({x}\right)+{sin}\left(\mathrm{2}{x}\right)= \\ $$$$\mathrm{6}{x}−\mathrm{8}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+….\right)+ \\ $$$$\left(\mathrm{2}{x}−\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{\mathrm{128}{x}^{\mathrm{7}} }{\mathrm{7}!}+…\right)= \\ $$$$\left(\mathrm{6}−\mathrm{8}+\mathrm{2}\right){x}+\left(\frac{\mathrm{8}}{\mathrm{3}!}−\frac{\mathrm{8}}{\mathrm{3}!}\right){x}^{\mathrm{3}} +\left(\frac{−\mathrm{8}}{\mathrm{5}!}+\frac{\mathrm{32}}{\mathrm{5}!}\right){x}^{\mathrm{5}} +{ax}^{\mathrm{7}} +…. \\ $$$$=\frac{\mathrm{24}}{\mathrm{5}!}{x}^{\mathrm{5}} +{ax}^{\mathrm{7}} +… \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{6}{x}−\mathrm{8}{sinx}+{sin}\mathrm{2}{x}}{{x}^{\mathrm{5}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{24}}{\mathrm{5}!}+{ax}^{\mathrm{2}} +…\right)= \\ $$$$\frac{\mathrm{24}}{\mathrm{5}!}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by MJS_new last updated on 23/Jan/23
$$\mathrm{but}\:\mathrm{it}\:\mathrm{says}\:“{without}\:\mathrm{series}''… \\ $$