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Question Number 144473 by SOMEDAVONG last updated on 25/Jun/21
L=limx→π38cos25x+2cosx−34cos25x+8cosx−5=?
Answered by mathmax by abdo last updated on 25/Jun/21
L=limx→π38(1+cos(10x)2)+2cosx−34(1+cos(10x)2)+8cosx−5=limx→π34+4cos(10x)+2cosx−32+2cos(10x)+8cosx−5=limx→π34cos(10x)+2cosx+12cos(10x)+8cosx−3=limx→π3−40sin(10x)−2sinx−20sin(10x)−8sinx(hospital)=40sin(10π3)+2sin(π3)20sin(10π3)+8sin(π3)=20sin(π+π3)+3210sin(π+π3)+4.32=−20.32+32−10.32+23=−103+32−53+23=−1932(−33)=196
Answered by EDWIN88 last updated on 26/Jun/21
L=limx→π/3−40sin10x−2sinx−20sin10x−8sinxL=limx→π/320sin10x+sinx10sin10x+4sinxL=−20sinπ3+sinπ3−10sinπ3+4sinπ3L=−20+1−10+4=196
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