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Question Number 111414 by bobhans last updated on 03/Sep/20
(1)limx→0sinx−ln(excosx)xsinx(2)limx→11−x+ln(x)1−2x−x2
Answered by john santu last updated on 03/Sep/20
(2)limx→1−1+1x−(2−2x22x−x2)=limx→11−xx.2x−x21−x=1
(1)limx→0cosx−(excosx−exsinxexcosx)sinx+xcosx=limx→0excos2x−excosx+exsinxexcosx(sinx+xcosx)=1×limx→0cos2x−cosx+sinxsinx+xcosx=limx→0(1−x22)2−(1−12x2)+x−x36x−x36+x(1−12x2)=limx→0x−12x2−x362x−13x3=limx→01−12x−x262−13x2=12.
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