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Question Number 159142 by tounghoungko last updated on 13/Nov/21
limx→−π4π8−2x.tanxsinx+cosx=?
Commented by cortano last updated on 14/Nov/21
limx→−π4πcosx−4xsinx82sin(x+π4)cosx=122limx→−π4πcosx−4xsinxsin(x+π4)=122limx→−π4−πsinx−4(sinx+xcosx)cos(x+π4)=122(π2+22+π21)=122(22+π2)=2+π2―
Answered by FongXD last updated on 13/Nov/21
L=limx→−π4π8(tanx+1)−(π8+2x)tanx2(sinπ4sinx+cosπ4cosx)⇔L=14limx→−π4πsin(x+π4)cosxcosπ4−(π+4x)tanxcos(x−π4)⇔L=14limx→−π4πsin(x+π4)−(π+4x)tanxcosxcosπ4sin(x+π4)×1cosxcosπ4⇔L=π2−12limx→−π4π+4xsin(x+π4)×tanxcosxcosπ4⇔L=π2+14limx→−π4π+4xsin(x+π4)putt=x+π4,ifx→−π4,⇒t→0weget:L=π2+14limt→04tsint=π2+14(4×1)=π2+1so.L=limx→−π4π8−2xtanxsinx+cosx=π2+1
Answered by tounghoungko last updated on 14/Nov/21
=12limx→−π4πcosx−4xsinx2sin(x+π4)=122limx→0πcos(x−π4)−(4x−π)sin(x−π4)sinx=122limx→0(π2−22x)sinx+22xcosxsinx=122limx→0(π2−22x+22cosx)=π2+2222=π2+1
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