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Question Number 82887 by M±th+et£s last updated on 25/Feb/20
findwithoutusingl′hopitallimx→0ln(1+cos(x))−ln(2)x
Commented by msup trace by abdo last updated on 25/Feb/20
wehavecosx∼1−x22⇒2+cosx∼2−x22⇒ln(1+cosx)∼ln(2)+ln(1−x24)∼ln(2)−x24⇒ln(1+cosx)−ln2x∼−x4→0(x→0)⇒limx→0ln(1+cosx)−ln2x=0
Answered by TANMAY PANACEA last updated on 25/Feb/20
limx→0ln(1+cosx2)xlimx→0ln(1+1+cosx2−1)(1+cosx2−1)×1+cosx2−1xt=(1+cosx2−1)whenx→0t→0limt→0ln(1+t)t×limx→0−2sinx2×sinx2x2×21×−1×0=0tlimx→0
Commented by M±th+et£s last updated on 25/Feb/20
thankyousir
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