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Question Number 116417 by bemath last updated on 03/Oct/20
limx→01+xsinx−cos2xtan2(x2)=?
Answered by Bird last updated on 03/Oct/20
f(x)=1+xsinx−cos(2x)tan2(x2)wehsve1+xsinx∼1+x2∼1+x22cos(2x)∼1−2x2∼1−x2tan2(x2)∼x24⇒f(x)∼1+x22−1+x2x24⇒f(x)∼4.32x2x2=6⇒limx→0f(x)=6
Answered by bobhans last updated on 04/Oct/20
limx→0(1+xsinx)−(1−2sin2x)tan2(12x)×limx→011+xsinx+cos2x=limx→0xsinx+2sin2xtan2(12x)×12=12×limx→0[xsinxx2+2sin2xx2][tan2(12x)x2]=12×3(14)=6.
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