Question Number 114696 by bemath last updated on 20/Sep/20
$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sinh}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\mathrm{2}{x}}{{x}^{\mathrm{5}} }\:=? \\ $$
Answered by bobhans last updated on 20/Sep/20
$${e}^{\mathrm{2}{x}} =\mathrm{1}+\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{3}}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{120}}+… \\ $$$${e}^{\mathrm{2}{x}} =\mathrm{1}−\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{3}}−\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{120}}+… \\ $$$$\mathrm{sinh}\:\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{x}+\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{64}{x}^{\mathrm{5}} }{\mathrm{120}}+…\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{x}+\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{120}}+… \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sinh}\:\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{x}}{{x}^{\mathrm{5}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{x}+\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{120}}\right)−\left(\mathrm{2}{x}−\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{120}}\right)}{{x}^{\mathrm{5}} }= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}}}{{x}^{\mathrm{5}} }\:=\:\mathrm{0}\: \\ $$