Question Number 185539 by mathlove last updated on 23/Jan/23
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}{e}^{\mathrm{3}{x}} −\mathrm{9}{e}^{\mathrm{2}{x}} +\mathrm{6}{x}+\mathrm{5}}{{x}^{\mathrm{3}} }=? \\ $$
Answered by Ar Brandon last updated on 23/Jan/23
$$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}{e}^{\mathrm{3}{x}} −\mathrm{9}{e}^{\mathrm{2}{x}} +\mathrm{6}{x}+\mathrm{5}}{{x}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\left(\mathrm{1}+\mathrm{3}{x}+\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{27}{x}^{\mathrm{3}} }{\mathrm{6}}\right)−\mathrm{9}\left(\mathrm{1}+\mathrm{2}{x}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{6}}\right)+\mathrm{6}{x}+\mathrm{5}}{{x}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{18}{x}^{\mathrm{3}} −\mathrm{12}{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} }=\mathrm{6} \\ $$
Commented by mathlove last updated on 23/Jan/23
$${thanks} \\ $$