Question Number 203247 by mathlove last updated on 13/Jan/24
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}\:{tan}\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+{x}\right)=? \\ $$
Answered by MM42 last updated on 13/Jan/24
$$={lim}_{{x}\rightarrow\mathrm{0}} \:−{xcot}\frac{\pi}{\mathrm{2}}{x}={lim}_{{x}\rightarrow\mathrm{0}} \:−\frac{\frac{\pi}{\mathrm{2}}{xcos}\frac{\pi}{\mathrm{2}}{x}}{{sin}\frac{\pi}{\mathrm{2}}{x}}×\frac{\mathrm{2}}{\pi} \\ $$$$=\:−\frac{\mathrm{2}}{\pi}\:\checkmark \\ $$$$ \\ $$
Commented by mathlove last updated on 13/Jan/24
$${way}\:{tan}\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+{x}\right)=−{cotx} \\ $$
Commented by MM42 last updated on 13/Jan/24
$${ok} \\ $$$$−{cot}\frac{\pi}{\mathrm{2}}{x} \\ $$
Answered by mr W last updated on 13/Jan/24
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}\:\mathrm{tan}\:\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+{x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}+\frac{\pi{x}}{\mathrm{2}}\right)}{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\frac{\pi{x}}{\mathrm{2}}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\pi{x}}{\mathrm{2}}\right)}{−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\pi{x}}{\mathrm{2}}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\:\mathrm{cos}\:\left(\frac{\pi{x}}{\mathrm{2}}\right)}{−\mathrm{sin}\:\left(\frac{\pi{x}}{\mathrm{2}}\right)} \\ $$$$=−\frac{\mathrm{2}}{\pi}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\frac{\pi{x}}{\mathrm{2}}\right)}{\mathrm{sin}\:\left(\frac{\pi{x}}{\mathrm{2}}\right)}×\mathrm{cos}\:\left(\frac{\pi{x}}{\mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{2}}{\pi} \\ $$
Commented by mathlove last updated on 13/Jan/24
$${thats}\:{right} \\ $$$${thanks}\:{sir} \\ $$