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Question Number 144473 by SOMEDAVONG last updated on 25/Jun/21

L=lim_(x→(π/3)) ((8cos^2 5x+2cosx−3)/(4cos^2 5x+8cosx−5))  =?

L=limxπ38cos25x+2cosx34cos25x+8cosx5=?

Answered by mathmax by abdo last updated on 25/Jun/21

L=lim_(x→(π/3))        ((8(((1+cos(10x))/2))+2cosx−3)/(4(((1+cos(10x))/2))+8cosx −5))  =lim_(x→(π/3))    ((4+4cos(10x)+2cosx−3)/(2+2cos(10x)+8cosx −5))  =lim_(x→(π/3))     ((4cos(10x)+2cosx+1)/(2cos(10x)+8cosx−3))  =lim_(x→(π/3))    ((−40sin(10x)−2sinx)/(−20sin(10x)−8sinx))  (hospital)  =((40sin(((10π)/3))+2sin((π/3)))/(20sin(10(π/3))+8sin((π/3))))=((20sin(π+(π/3))+((√3)/2))/(10sin(π+(π/3))+4.((√3)/2)))  =((−20.((√3)/2)+((√3)/2))/(−10.((√3)/2)+2(√3)))=((−10(√3)+((√3)/2))/(−5(√3)+2(√3)))=((−19(√3))/(2(−3(√3)))) =((19)/6)

L=limxπ38(1+cos(10x)2)+2cosx34(1+cos(10x)2)+8cosx5=limxπ34+4cos(10x)+2cosx32+2cos(10x)+8cosx5=limxπ34cos(10x)+2cosx+12cos(10x)+8cosx3=limxπ340sin(10x)2sinx20sin(10x)8sinx(hospital)=40sin(10π3)+2sin(π3)20sin(10π3)+8sin(π3)=20sin(π+π3)+3210sin(π+π3)+4.32=20.32+3210.32+23=103+3253+23=1932(33)=196

Answered by EDWIN88 last updated on 26/Jun/21

L=lim_(x→π/3) ((−40sin 10x−2sin x)/(−20sin 10x−8sin x))  L=lim_(x→π/3) ((20sin 10x+sin x)/(10sin 10x+4sin x))  L=((−20sin (π/3)+sin (π/3))/(−10sin (π/3)+4sin (π/3)))  L=((−20+1)/(−10+4))=((19)/6)

L=limxπ/340sin10x2sinx20sin10x8sinxL=limxπ/320sin10x+sinx10sin10x+4sinxL=20sinπ3+sinπ310sinπ3+4sinπ3L=20+110+4=196

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