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Question Number 134378 by mohammad17 last updated on 02/Mar/21

lim_(x→3) ((cos(((3π)/(2x))))/(x−3))

$${lim}_{{x}\rightarrow\mathrm{3}} \frac{{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{2}{x}}\right)}{{x}−\mathrm{3}} \\ $$

Answered by bramlexs22 last updated on 02/Mar/21

lim_(x−3→0) ((cos (((3π)/(2x))))/(x−3)) let x−3=w ⇒ (1/x)=(1/(w+3)) lim_(w→0) ((cos (((3π)/(2(w+3)))))/w) = lim_(w→0) ((((3π)/(2(w+3)^2 )).sin (((3π)/(2(w+3)))))/1) = (π/6)

$$\underset{{x}−\mathrm{3}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2x}}\right)}{\mathrm{x}−\mathrm{3}} \\ $$$$\mathrm{let}\:\mathrm{x}−\mathrm{3}=\mathrm{w}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{x}}=\frac{\mathrm{1}}{\mathrm{w}+\mathrm{3}} \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}\left(\mathrm{w}+\mathrm{3}\right)}\right)}{\mathrm{w}}\:= \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{3}\pi}{\mathrm{2}\left(\mathrm{w}+\mathrm{3}\right)^{\mathrm{2}} }.\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}\left(\mathrm{w}+\mathrm{3}\right)}\right)}{\mathrm{1}}\:=\:\frac{\pi}{\mathrm{6}} \\ $$

Commented by mr W last updated on 03/Mar/21

((3π)/2)×(1/3^2 )

$$\frac{\mathrm{3}\pi}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} } \\ $$

Commented by bramlexs22 last updated on 02/Mar/21

(d/dw) [cos (((3π)/(2(w+3))))]= (d/dw) [cos ((3π)/2)(w+3)^(−1) ]= −((3π)/2).(w+3)^(−2) [−sin (((3π)/(2(w+3))))] lim_(w→0) ((3π)/(2(w+3)^2 )).sin (((3π)/(2(w+3))))= ((3π)/(18))×1=(π/6)

$$\frac{\mathrm{d}}{\mathrm{dw}}\:\left[\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}\left(\mathrm{w}+\mathrm{3}\right)}\right)\right]= \\ $$$$\frac{\mathrm{d}}{\mathrm{dw}}\:\left[\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{2}}\left(\mathrm{w}+\mathrm{3}\right)^{−\mathrm{1}} \:\right]= \\ $$$$−\frac{\mathrm{3}\pi}{\mathrm{2}}.\left(\mathrm{w}+\mathrm{3}\right)^{−\mathrm{2}} \:\left[−\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}\left(\mathrm{w}+\mathrm{3}\right)}\right)\right] \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{3}\pi}{\mathrm{2}\left(\mathrm{w}+\mathrm{3}\right)^{\mathrm{2}} }.\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}\left(\mathrm{w}+\mathrm{3}\right)}\right)= \\ $$$$\frac{\mathrm{3}\pi}{\mathrm{18}}×\mathrm{1}=\frac{\pi}{\mathrm{6}} \\ $$

Answered by mathmax by abdo last updated on 02/Mar/21

f(x)=((cos(((3π)/(2x))))/(x−3)) changement ((3π)/(2x))=(π/2)+t x→3 ⇒t→0 and ((3π)/(2x))=((2t+π)/2) ⇒((3π)/x)=2t+π ⇒(x/(3π))=(1/(2t+π)) ⇒ x=((3π)/(2t+π)) ⇒x−3=((3π)/(2t+π))−3 =((3π−6t−3π)/(2t+π)) =((−6t)/(2t+π)) ⇒ f(x)=g(t)=((cos((π/2)+t))/((−6t)/(2t+π))) =(((2t+π)sint)/(6t)) ⇒g(t)∼((2t+π)/6) →(π/6) ⇒ lim_(x→3) f(x)=(π/6)

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{2x}}\right)}{\mathrm{x}−\mathrm{3}}\:\:\mathrm{changement}\:\frac{\mathrm{3}\pi}{\mathrm{2x}}=\frac{\pi}{\mathrm{2}}+\mathrm{t} \\ $$$$\mathrm{x}\rightarrow\mathrm{3}\:\Rightarrow\mathrm{t}\rightarrow\mathrm{0}\:\mathrm{and}\:\frac{\mathrm{3}\pi}{\mathrm{2x}}=\frac{\mathrm{2t}+\pi}{\mathrm{2}}\:\Rightarrow\frac{\mathrm{3}\pi}{\mathrm{x}}=\mathrm{2t}+\pi\:\Rightarrow\frac{\mathrm{x}}{\mathrm{3}\pi}=\frac{\mathrm{1}}{\mathrm{2t}+\pi}\:\Rightarrow \\ $$$$\mathrm{x}=\frac{\mathrm{3}\pi}{\mathrm{2t}+\pi}\:\Rightarrow\mathrm{x}−\mathrm{3}=\frac{\mathrm{3}\pi}{\mathrm{2t}+\pi}−\mathrm{3}\:=\frac{\mathrm{3}\pi−\mathrm{6t}−\mathrm{3}\pi}{\mathrm{2t}+\pi}\:=\frac{−\mathrm{6t}}{\mathrm{2t}+\pi}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{t}\right)=\frac{\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}+\mathrm{t}\right)}{\frac{−\mathrm{6t}}{\mathrm{2t}+\pi}}\:=\frac{\left(\mathrm{2t}+\pi\right)\mathrm{sint}}{\mathrm{6t}}\:\Rightarrow\mathrm{g}\left(\mathrm{t}\right)\sim\frac{\mathrm{2t}+\pi}{\mathrm{6}}\:\rightarrow\frac{\pi}{\mathrm{6}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{3}} \mathrm{f}\left(\mathrm{x}\right)=\frac{\pi}{\mathrm{6}} \\ $$$$ \\ $$

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