lim tan x-sinx/x3 x--0 Share with your friends Share 1 Anuradha Sharma answered this Given , limx→0 tanx-sinxx3After putting limit we are getting 00 form so using L'hospital rule , limx→0 ddxtanx-sinxddxx3=limx→0 sec2x-cos x3x2Again using L'hospital rule we get, limx→0 ddxsec2x-cos xddx3x2=limx→0 2secx.secxtanx+sinx6x=limx→0 2sinxcos3x+sinx6x=limx→02cos3xsinxx+sinxx6Now as we know that limx→0sinxx=1 so putting the limit we get, =2+16=12 10 View Full Answer Isha Chaudhari answered this r u sure about d question? 0