Question Number 106771 by bemath last updated on 07/Aug/20
![@bemath@ lim_(x→π) [((4(x−π) cos^2 x)/(π(π−2x) cos (x−(π/2))))]= ?](https://www.tinkutara.com/question/Q106771.png)
$$\:\:\:\:\:\:\:@\mathrm{bemath}@ \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\left[\frac{\mathrm{4}\left(\mathrm{x}−\pi\right)\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\pi\left(\pi−\mathrm{2x}\right)\:\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{2}}\right)}\right]=\:? \\ $$
Commented by bemath last updated on 07/Aug/20
Answered by john santu last updated on 07/Aug/20
![^(@JS@) lim_(x→π) [((cos^2 x)/(π(π−2x)))] × lim_(x→π) ((4(x−π))/(cos (x−(π/2))))= (((−1)^2 )/(−π^2 )) × lim_(x→π) (4/(−sin (x−(π/2)))) = (1/π^2 ) × 4 = (4/π^2 ) . (by L′Hopital′s rule)](https://www.tinkutara.com/question/Q106773.png)
$$\:\:\:\:\:\:\:\:^{@\mathrm{JS}@} \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\left[\frac{\mathrm{cos}\:^{\mathrm{2}} \:\mathrm{x}}{\pi\left(\pi−\mathrm{2x}\right)}\right]\:×\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{4}\left(\mathrm{x}−\pi\right)}{\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{2}}\right)}= \\ $$$$\frac{\left(−\mathrm{1}\right)^{\mathrm{2}} }{−\pi^{\mathrm{2}} }\:×\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{4}}{−\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{2}}\right)}\:= \\ $$$$\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\:×\:\mathrm{4}\:=\:\frac{\mathrm{4}}{\pi^{\mathrm{2}} }\:.\:\:\left(\mathrm{by}\:\mathrm{L}'\mathrm{Hopital}'\mathrm{s}\:\mathrm{rule}\right) \\ $$
Answered by Dwaipayan Shikari last updated on 07/Aug/20
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\mathrm{4}{cos}^{\mathrm{2}} {x}}{\pi\left(\pi−\mathrm{2}{x}\right)}\:\frac{\left({x}−\pi\right)}{{cos}\left({x}−\frac{\pi}{\mathrm{2}}\right)} \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}−\frac{\mathrm{4}}{\pi^{\mathrm{2}} }\:\frac{{x}−\pi}{{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{4}}{\pi^{\mathrm{2}} }.\:\frac{{x}−\pi}{{sinx}} \\ $$$$=\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\mathrm{4}}{\pi^{\mathrm{2}} }.\frac{\pi−{x}}{{sin}\left(\pi−{x}\right)}=\frac{\mathrm{4}}{\pi^{\mathrm{2}} } \\ $$