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Question Number 82887 by M±th+et£s last updated on 25/Feb/20

find without using l′hopital  lim_(x→0) ((ln(1+cos(x))−ln(2))/x)

findwithoutusinglhopitallimx0ln(1+cos(x))ln(2)x

Commented by msup trace by abdo last updated on 25/Feb/20

we have cosx ∼1−(x^2 /2) ⇒  2+cosx ∼2−(x^2 /2) ⇒  ln(1+cosx)∼ln(2)+ln(1−(x^2 /4))  ∼ln(2)−(x^2 /4) ⇒((ln(1+cosx)−ln2)/x)  ∼−(x/4)→0 (x→0) ⇒  lim_(x→0)   ((ln(1+cosx)−ln2)/x)=0

wehavecosx1x222+cosx2x22ln(1+cosx)ln(2)+ln(1x24)ln(2)x24ln(1+cosx)ln2xx40(x0)limx0ln(1+cosx)ln2x=0

Answered by TANMAY PANACEA last updated on 25/Feb/20

lim_(x→0)  ((ln(((1+cosx)/2)))/x)  lim_(x→0) ((ln(1+((1+cosx)/2)−1))/((((1+cosx)/2)−1)))×((((1+cosx)/2)−1)/x)  t=(((1+cosx)/2)−1)  when x→0   t→0  lim_(t→0) ((ln(1+t))/t)×lim_(x→0) ((−2sin(x/2)×sin(x/2))/((x/2)×2))  1×−1×0=0          t  lim_(x→0)

limx0ln(1+cosx2)xlimx0ln(1+1+cosx21)(1+cosx21)×1+cosx21xt=(1+cosx21)whenx0t0limt0ln(1+t)t×limx02sinx2×sinx2x2×21×1×0=0tlimx0

Commented by M±th+et£s last updated on 25/Feb/20

thank you sir

thankyousir

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