Question Number 76248 by john santuy last updated on 25/Dec/19
$$\overset{{lim}} {{x}}\rightarrow\overset{\:\:\:\:\left(\frac{{sin}\mathrm{3}{x}}{\mathrm{2}{x}}\right)^{\frac{\mathrm{2}}{\mathrm{5}{x}+\mathrm{1}}} } {\mathrm{0}}=\:?\:\: \\ $$
Commented by Mikael_786 last updated on 25/Dec/19
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\mathrm{3}{x}}{\mathrm{2}{x}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{3}{sin}\mathrm{3}{x}}{\mathrm{3}.\mathrm{2}{x}}=\frac{\mathrm{3}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\mathrm{3}{x}}{\mathrm{3}{x}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{2}}{\mathrm{5}{x}+\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{5}×\mathrm{0}+\mathrm{1}}=\mathrm{2} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{{sin}\mathrm{3}{x}}{\mathrm{2}{x}}\right)^{\frac{\mathrm{2}}{\mathrm{5}{x}+\mathrm{1}}} =\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}} \\ $$
Commented by turbo msup by abdo last updated on 25/Dec/19
$${let}\:{f}\left({x}\right)=\left(\frac{{sin}\left(\mathrm{3}{x}\right)}{\mathrm{2}{x}}\right)^{\frac{\mathrm{2}}{\mathrm{5}{x}+\mathrm{1}}} \\ $$$$\Rightarrow{f}\left({x}\right)={e}^{\frac{\mathrm{2}}{\mathrm{5}{x}+\mathrm{1}}{ln}\left(\frac{{sin}\left(\mathrm{3}{x}\right)}{\mathrm{2}{x}}\right)} \:\:\:{we}\:{have} \\ $$$$\frac{{sin}\left(\mathrm{3}{x}\right)}{\mathrm{2}{x}}\:\sim\:\frac{\mathrm{3}{x}}{\mathrm{2}{x}}\:=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right)=\:{e}^{\mathrm{2}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \:=\frac{\mathrm{9}}{\mathrm{4}} \\ $$
Commented by john santuy last updated on 26/Dec/19
$${thanks}\:{sir} \\ $$