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

lim_(x→3) ((cos(((3π)/(2x))))/(x−3))

limx3cos(3π2x)x3

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)

limx30cos(3π2x)x3letx3=w1x=1w+3limw0cos(3π2(w+3))w=limw03π2(w+3)2.sin(3π2(w+3))1=π6

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

((3π)/2)×(1/3^2 )

3π2×132

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)

ddw[cos(3π2(w+3))]=ddw[cos3π2(w+3)1]=3π2.(w+3)2[sin(3π2(w+3))]limw03π2(w+3)2.sin(3π2(w+3))=3π18×1=π6

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)

f(x)=cos(3π2x)x3changement3π2x=π2+tx3t0and3π2x=2t+π23πx=2t+πx3π=12t+πx=3π2t+πx3=3π2t+π3=3π6t3π2t+π=6t2t+πf(x)=g(t)=cos(π2+t)6t2t+π=(2t+π)sint6tg(t)2t+π6π6limx3f(x)=π6

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