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Question Number 142459 by bramlexs22 last updated on 01/Jun/21
−−−−−−−−−−− lim_(x→(π/2)) ((x/(cot x))−(π/(2cos x)))=? ___________________
$$\:\:\:\:\:\:−−−−−−−−−−− \\ $$$$\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left(\frac{{x}}{\mathrm{cot}\:{x}}−\frac{\pi}{\mathrm{2cos}\:{x}}\right)=? \\ $$$$\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$
Answered by benjo_mathlover last updated on 01/Jun/21
lim_(x→(π/2)) (((x sin x)/(cos x))−(π/(2 cos x))) = lim_(x→(π/2)) (((2x sin x−π)/(2 cos x))) = lim_(x→(π/2)) (((2sin x +2x cos x)/(−2sin x))) = −1
$$\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left(\frac{{x}\:\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}−\frac{\pi}{\mathrm{2}\:\mathrm{cos}\:{x}}\right) \\ $$$$=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left(\frac{\mathrm{2}{x}\:\mathrm{sin}\:{x}−\pi}{\mathrm{2}\:\mathrm{cos}\:{x}}\right) \\ $$$$=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left(\frac{\mathrm{2sin}\:{x}\:+\mathrm{2}{x}\:\mathrm{cos}\:{x}}{−\mathrm{2sin}\:{x}}\right) \\ $$$$=\:−\mathrm{1} \\ $$
Answered by mathmax by abdo last updated on 02/Jun/21
f(x)=(x/(cotanx))−(π/(2cosx)) ⇒f(x)=((xsinx)/(cosx))−(π/(2cosx))=((2xsinx−π)/(2cosx)) changement x=(π/2)−t give f(x)=f((π/2)−t)=((2((π/2)−t)cost−π)/(2sint)) =(((π−2t)cost−π)/(2sint)) ∼(((π−2t)(1−(t^2 /2))−π)/(2t)) =((π−(π/(2 ))t^2 −2t+t^3 −π)/(2t))=−(π/4)t−1+(t^2 /2) →−1 (t→0) ⇒ lim_(x→(π/2)) f(x)=−1
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{cotanx}}−\frac{\pi}{\mathrm{2cosx}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{xsinx}}{\mathrm{cosx}}−\frac{\pi}{\mathrm{2cosx}}=\frac{\mathrm{2xsinx}−\pi}{\mathrm{2cosx}} \\ $$$$\mathrm{changement}\:\mathrm{x}=\frac{\pi}{\mathrm{2}}−\mathrm{t}\:\mathrm{give}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)=\frac{\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)\mathrm{cost}−\pi}{\mathrm{2sint}} \\ $$$$=\frac{\left(\pi−\mathrm{2t}\right)\mathrm{cost}−\pi}{\mathrm{2sint}}\:\sim\frac{\left(\pi−\mathrm{2t}\right)\left(\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\right)−\pi}{\mathrm{2t}} \\ $$$$=\frac{\pi−\frac{\pi}{\mathrm{2}\:}\mathrm{t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{t}^{\mathrm{3}} −\pi}{\mathrm{2t}}=−\frac{\pi}{\mathrm{4}}\mathrm{t}−\mathrm{1}+\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:\rightarrow−\mathrm{1}\:\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \mathrm{f}\left(\mathrm{x}\right)=−\mathrm{1} \\ $$

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