Question Number 195325 by mathlove last updated on 30/Jul/23
$${prove}\:{that} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1}}{{x}−\frac{\pi}{\mathrm{2}}}=\mathrm{1} \\ $$
Answered by BaliramKumar last updated on 30/Jul/23
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\frac{\mathrm{d}}{\mathrm{dx}}\left({tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1}\right)}{\frac{\mathrm{d}}{\mathrm{dx}}\left({x}−\frac{\pi}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{sec}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\centerdot\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:=\:\mathrm{1} \\ $$
Answered by som(math1967) last updated on 30/Jul/23
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{lim}}\:\frac{{sin}\frac{{x}}{\mathrm{2}}−{cos}\frac{{x}}{\mathrm{2}}}{{cos}\frac{{x}}{\mathrm{2}}\left({x}−\frac{\pi}{\mathrm{2}}\right)} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{lim}}\:\frac{\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sin}\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{cos}\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{cos}\frac{{x}}{\mathrm{2}}\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{lim}}\:\frac{\sqrt{\mathrm{2}}{sin}\left(\frac{{x}}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{lim}}\:\frac{{sin}\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)}{\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$${let}\:\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)={t} \\ $$$$\:{x}\rightarrow\frac{\pi}{\mathrm{2}}\Rightarrow\frac{{x}}{\mathrm{2}}\rightarrow\frac{\pi}{\mathrm{4}}\:\therefore\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\rightarrow\mathrm{0} \\ $$$$\therefore\:\underset{{t}\rightarrow\mathrm{0}} {{lim}}\frac{{sint}}{{t}}=\mathrm{1} \\ $$
Commented by mathlove last updated on 30/Jul/23
$${tnks} \\ $$