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lim-x-pi-2-cos-2x-sin-3pi-2-x-




Question Number 130406 by john_santu last updated on 25/Jan/21
lim_(x→π/2)  cos (2x+sin (((3π)/2)+x)=?
limxπ/2cos(2x+sin(3π2+x)=?
Commented by Dwaipayan Shikari last updated on 25/Jan/21
−1
1
Answered by EDWIN88 last updated on 25/Jan/21
 lim_(x→a)  f(g(x))= f(lim_(x→a)  g(x))   lim_(x→π/2)  cos (2x+sin (((3π)/2)+x))=   cos (lim_(x→π/2)  (2x+sin (((3π)/2)+x))=   cos (π+sin 2π) = cos π = −1
limxaf(g(x))=f(limxag(x))limxπ/2cos(2x+sin(3π2+x))=cos(limxπ/2(2x+sin(3π2+x))=cos(π+sin2π)=cosπ=1
Answered by mathmax by abdo last updated on 25/Jan/21
f(x)=cos(2x+sin(((3π)/2)+x))⇒f(x)=cos(2x+sin(π+(π/2)+x))  =cos(2x−cosx)  ⇒lim_(x→(π/2)) f(x)=cos(π−0)=−1
f(x)=cos(2x+sin(3π2+x))f(x)=cos(2x+sin(π+π2+x))=cos(2xcosx)limxπ2f(x)=cos(π0)=1

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