Question Number 197281 by cortano12 last updated on 12/Sep/23
$$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{x}+\mathrm{2x}^{\mathrm{5}} }{\mathrm{3x}^{\mathrm{3}} }\:=? \\ $$
Answered by MM42 last updated on 12/Sep/23
$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{5}} }{\mathrm{3}{x}^{\mathrm{3}} }\:=−\frac{\mathrm{1}}{\mathrm{18}}\:\checkmark \\ $$
Answered by tri26112004 last updated on 13/Sep/23
$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{cos}\:{x}\:+\:\mathrm{10}{x}^{\mathrm{4}} −\mathrm{1}}{\mathrm{9}{x}^{\mathrm{2}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−{sin}\:{x}\:+\:\mathrm{40}{x}^{\mathrm{3}} }{\mathrm{18}{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−{cos}\:{x}\:+\:\mathrm{120}{x}^{\mathrm{2}} }{\mathrm{18}} \\ $$$$\:=\:−\:\frac{\mathrm{1}}{\mathrm{18}} \\ $$