Question Number 119074 by benjo_mathlover last updated on 22/Oct/20
$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left({x}−\mathrm{1}\right)\:\mathrm{tan}\:\left(\frac{\pi{x}}{\mathrm{2}}\right)\: \\ $$
Answered by Dwaipayan Shikari last updated on 22/Oct/20
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left({x}−\mathrm{1}\right){cot}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}{x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}−\left(\mathrm{1}−{x}\right)\frac{{cos}\left(\mathrm{1}−{x}\right)\frac{\pi}{\mathrm{2}}}{{sin}\left(\mathrm{1}−{x}\right)\frac{\pi}{\mathrm{2}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}−\left(\mathrm{1}−{x}\right)\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)\frac{\pi}{\mathrm{2}}}=−\frac{\mathrm{2}}{\pi}\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{sinx}={x} \\ $$
Answered by bemath last updated on 22/Oct/20
![let x = 1+m ∧ m→0 lim_(m→0) (m/(cot ((((1+m)π)/2)))) = lim_(m→0) [ (m/(−tan (((mπ)/2)))) ] = −(2/π)](https://www.tinkutara.com/question/Q119075.png)
$$\:{let}\:{x}\:=\:\mathrm{1}+{m}\:\wedge\:{m}\rightarrow\mathrm{0} \\ $$$$\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{m}}{\mathrm{cot}\:\left(\frac{\left(\mathrm{1}+{m}\right)\pi}{\mathrm{2}}\right)}\:=\:\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\:\frac{{m}}{−\mathrm{tan}\:\left(\frac{{m}\pi}{\mathrm{2}}\right)}\:\right] \\ $$$$=\:−\frac{\mathrm{2}}{\pi} \\ $$
Answered by 1549442205PVT last updated on 22/Oct/20
![Put x−1=t⇒(x→1∼t→0∼πt/2→0) I= lim (x−1) tan (((πx)/2)) =lim_(t→0) [ttan((π(1+t))/2)] =lim_(t→0) (−tcot((πt)/2))=lim_(t→0) (−(1/π))(((πt)/(sin((πt)/2)))×cos((πt)/2)) =lim_(t→0) (−(1/π))×(( 2)/((sin(πt/2))/((πt/2))))×cos(πt/2)= −(2/π)×1=−(2/π)](https://www.tinkutara.com/question/Q119087.png)
$$\mathrm{Put}\:\mathrm{x}−\mathrm{1}=\mathrm{t}\Rightarrow\left(\mathrm{x}\rightarrow\mathrm{1}\sim\mathrm{t}\rightarrow\mathrm{0}\sim\pi\mathrm{t}/\mathrm{2}\rightarrow\mathrm{0}\right) \\ $$$$\mathrm{I}=\:\mathrm{lim}\:\left({x}−\mathrm{1}\right)\:\mathrm{tan}\:\left(\frac{\pi{x}}{\mathrm{2}}\right)\:=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{ttan}\frac{\pi\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{2}}\right] \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(−\mathrm{tcot}\frac{\pi\mathrm{t}}{\mathrm{2}}\right)=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(−\frac{\mathrm{1}}{\pi}\right)\left(\frac{\pi\mathrm{t}}{\mathrm{sin}\frac{\pi\mathrm{t}}{\mathrm{2}}}×\mathrm{cos}\frac{\pi\mathrm{t}}{\mathrm{2}}\right) \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(−\frac{\mathrm{1}}{\pi}\right)×\frac{\:\mathrm{2}}{\frac{\mathrm{sin}\left(\pi\mathrm{t}/\mathrm{2}\right)}{\left(\pi\mathrm{t}/\mathrm{2}\right)}}×\mathrm{cos}\left(\pi\mathrm{t}/\mathrm{2}\right)= \\ $$$$−\frac{\mathrm{2}}{\pi}×\mathrm{1}=−\frac{\mathrm{2}}{\pi} \\ $$
Answered by malwan last updated on 22/Oct/20
$$\underset{{x}\rightarrow\mathrm{1}} {{lim}}\:\left({x}−\mathrm{1}\right){cot}\left(\frac{\pi}{\mathrm{2}}\:−\:\frac{\pi{x}}{\mathrm{2}}\right) \\ $$$$=\:\underset{{x}\rightarrow\mathrm{1}} {{lim}}\:\left({x}−\mathrm{1}\right){cot}\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{x}\right) \\ $$$$=\:\underset{{x}\rightarrow\mathrm{1}} {{lim}}\:\frac{\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{x}\right)×\left(−\mathrm{1}\right)}{\frac{\pi}{\mathrm{2}}{tan}\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{x}\right)}\:=\:−\frac{\mathrm{2}}{\pi} \\ $$