Question Number 78325 by jagoll last updated on 16/Jan/20
$$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)^{\mathrm{3}{x}−\mathrm{1}} \\ $$
Commented by john santu last updated on 16/Jan/20
$${consider}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\left(\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{5}{x}}\right)}\right)^{\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{5}{x}}} ={e} \\ $$
Commented by john santu last updated on 16/Jan/20
$${by}\:{short}\:{cut}\: \\ $$$${e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}{x}−\mathrm{1}\right)\left(\frac{\mathrm{5}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)} =\:{e}^{\mathrm{3}} \\ $$
Commented by mathmax by abdo last updated on 16/Jan/20
$${let}\:{A}\left({x}\right)=\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)^{\mathrm{3}{x}−\mathrm{1}} \:\Rightarrow{A}\left({x}\right)={e}^{\left(\mathrm{3}{x}−\mathrm{1}\right){ln}\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)} \\ $$$${we}\:{have}\:{ln}\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)\:={ln}\left(\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)={ln}\left(\mathrm{1}+\frac{\mathrm{5}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right) \\ $$$$\sim\frac{\mathrm{5}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\:\sim\frac{\mathrm{1}}{{x}}\left({x}\rightarrow+\infty\right)\:\Rightarrow{A}\left({x}\right)\sim{e}^{\frac{\mathrm{3}{x}−\mathrm{1}}{{x}}} \:={e}^{\mathrm{3}} .{e}^{−\frac{\mathrm{1}}{{x}}} \:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} \:{A}\left({x}\right)={e}^{\mathrm{3}} \\ $$