Menu Close

lim-x-pi-2-1-sin-pi-2x-5-sin-x-pi-2-ln-4-cos-pi-2x-1-pi-2-x-2-

Tinku Tara 0 Comments
Pin




Question Number 170018 by cortano1 last updated on 14/May/22
lim_(x→(π/2)) (1+sin (π−2x))^(5/(sin (x−(π/2)))) .ln (4.((cos (π−2x)−1)/(((π/2)−x)^2 )))=?
$$\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{sin}\:\left(\pi−\mathrm{2}{x}\right)\right)^{\frac{\mathrm{5}}{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{2}}\right)}} .\mathrm{ln}\:\left(\mathrm{4}.\frac{\mathrm{cos}\:\left(\pi−\mathrm{2}{x}\right)−\mathrm{1}}{\left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}} }\right)=? \\ $$
Answered by Mathspace last updated on 14/May/22
u(x)=(1+sin(π−2x))^(5/(sin(x−(π/2)))) and v(x)=ln(4.((cos(π−2x)−1)/(((π/2)−x)^2 )) ⇒u(x)=(1+sin(2x))^(5/(−cosx)) =e^(−(5/(cosx))ln(1+sin(2x)) ch.x−(π/2)=t give u(x)=u((π/2)+t)=e^(−(5/(−sint))ln(1−sin(2t))) sint∼t and sin(2t)∼2t ⇒ ln(1−sin(2t))∼ln(1−2t)∼−2t⇒ (5/(sint))ln(1−sin(2t)∼(5/t)(−2t)=−10 ⇒ lim u(x)=e^(−10) v(x)=ln(4.((−cos(2x)−1)/(((π/2)−x)^2 )))(x−(π/2)=t) =ln(4×((−cos(2((π/2)+t)−1)/t^2 )) =ln(−4.((−cos(2t)+1)/t^2 )) =ln((4/t^2 )(cos(2t)−1)) but cos(2t)−1<0 error at the question!... cos(2t)∼1−2t^2 ⇒1−cos(2t)
$${u}\left({x}\right)=\left(\mathrm{1}+{sin}\left(\pi−\mathrm{2}{x}\right)\right)^{\frac{\mathrm{5}}{{sin}\left({x}−\frac{\pi}{\mathrm{2}}\right)}} \\ $$$${and}\:{v}\left({x}\right)={ln}\left(\mathrm{4}.\frac{{cos}\left(\pi−\mathrm{2}{x}\right)−\mathrm{1}}{\left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}} }\right. \\ $$$$\Rightarrow{u}\left({x}\right)=\left(\mathrm{1}+{sin}\left(\mathrm{2}{x}\right)\right)^{\frac{\mathrm{5}}{−{cosx}}} \\ $$$$={e}^{−\frac{\mathrm{5}}{{cosx}}{ln}\left(\mathrm{1}+{sin}\left(\mathrm{2}{x}\right)\right.} \\ $$$${ch}.{x}−\frac{\pi}{\mathrm{2}}={t}\:{give} \\ $$$${u}\left({x}\right)={u}\left(\frac{\pi}{\mathrm{2}}+{t}\right)={e}^{−\frac{\mathrm{5}}{−{sint}}{ln}\left(\mathrm{1}−{sin}\left(\mathrm{2}{t}\right)\right)} \\ $$$${sint}\sim{t}\:\:{and}\:{sin}\left(\mathrm{2}{t}\right)\sim\mathrm{2}{t}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{sin}\left(\mathrm{2}{t}\right)\right)\sim{ln}\left(\mathrm{1}−\mathrm{2}{t}\right)\sim−\mathrm{2}{t}\Rightarrow \\ $$$$\frac{\mathrm{5}}{{sint}}{ln}\left(\mathrm{1}−{sin}\left(\mathrm{2}{t}\right)\sim\frac{\mathrm{5}}{{t}}\left(−\mathrm{2}{t}\right)=−\mathrm{10}\:\Rightarrow\right. \\ $$$${lim}\:{u}\left({x}\right)={e}^{−\mathrm{10}} \\ $$$${v}\left({x}\right)={ln}\left(\mathrm{4}.\frac{−{cos}\left(\mathrm{2}{x}\right)−\mathrm{1}}{\left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}} }\right)\left({x}−\frac{\pi}{\mathrm{2}}={t}\right) \\ $$$$={ln}\left(\mathrm{4}×\frac{−{cos}\left(\mathrm{2}\left(\frac{\pi}{\mathrm{2}}+{t}\right)−\mathrm{1}\right.}{{t}^{\mathrm{2}} }\right) \\ $$$$={ln}\left(−\mathrm{4}.\frac{−{cos}\left(\mathrm{2}{t}\right)+\mathrm{1}}{{t}^{\mathrm{2}} }\right) \\ $$$$={ln}\left(\frac{\mathrm{4}}{{t}^{\mathrm{2}} }\left({cos}\left(\mathrm{2}{t}\right)−\mathrm{1}\right)\right) \\ $$$${but}\:{cos}\left(\mathrm{2}{t}\right)−\mathrm{1}<\mathrm{0}\:\:{error}\:{at}\:{the}\:{question}!… \\ $$$${cos}\left(\mathrm{2}{t}\right)\sim\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \:\Rightarrow\mathrm{1}−{cos}\left(\mathrm{2}{t}\right) \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *