How do you find the limit of #(costheta-1)/sintheta# as #theta->0#?

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Answer 1 · 2016-10-21T10:11:28

#Lt_(theta->0)(costheta-1)/sintheta=0#

#Lt_(theta->0)(costheta-1)/sintheta#

= #Lt_(theta->0)(costheta-1)/sinthetaxx(costheta+1)/(costheta+1)#

= #Lt_(theta->0)(cos^2theta-1)/(sinthetaxx(costheta+1))#

= #Lt_(theta->0)(-sin^2theta)/(sinthetaxx(costheta+1))#

= #Lt_(theta->0)(-sintheta)/(costheta+1)#

= #(-sin0)/(cos0+1)#

= #-0/2=0#
graph{(cosx-1)/sinx [-10, 10, -5, 5]}