Question Number 40092 by maxmathsup by imad last updated on 15/Jul/18
$${calculate}\:{lim}_{{x}\rightarrow+\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{{x}}\:{tan}\left(\frac{\pi{x}}{\mathrm{2}{x}+\mathrm{3}}\right) \\ $$
Commented by math khazana by abdo last updated on 26/Jul/18
$${we}\:{have}\:\frac{\pi{x}}{\mathrm{2}{x}+\mathrm{3}}\:=\frac{\pi{x}}{\mathrm{2}{x}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}{x}}\right)}\:=\frac{\pi}{\mathrm{2}}\:.\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}{x}}} \\ $$$$\sim\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}{x}}\:+\mathrm{0}\left(\frac{\mathrm{1}}{{x}}\right)\right)\:\left({x}\rightarrow+\infty\right)\Rightarrow \\ $$$${tan}\left(\frac{\pi{x}}{\mathrm{2}{x}+\mathrm{3}}\right)\:\sim{tan}\left(\frac{\pi}{\mathrm{2}}\:−\frac{\mathrm{3}\pi}{\mathrm{4}{x}}\right)\:\sim\:\frac{\mathrm{1}}{{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{4}{x}}\right)}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{x}}{tan}\left(\frac{\pi{x}}{\mathrm{2}{x}+\mathrm{3}}\right)\:\sim\:\:\frac{\mathrm{1}}{{x}\:{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{4}{x}}\right)}\:{changement}\: \\ $$$$\frac{\mathrm{3}\pi}{\mathrm{4}{x}}\:={t}\:{give}\mathrm{4}{tx}=\mathrm{3}\pi\:\:\Rightarrow\:{lim}_{{x}\rightarrow+\infty} \:\frac{\mathrm{1}}{{xtan}\left(\frac{\mathrm{3}\pi}{\mathrm{4}{x}}\right)} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{3}\pi}{\mathrm{4}{t}}{tan}\left({t}\right)}\:=\frac{\mathrm{4}}{\mathrm{3}\pi}{lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{\mathrm{1}}{\frac{{tant}}{{t}}}\:=\frac{\mathrm{4}}{\mathrm{3}\pi} \\ $$$${because}\:{lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\:\frac{{tant}}{{t}}\:=\mathrm{1} \\ $$