How do you find #lim_(theta->0)# #((1-cos^2 theta)/theta ^2)#?

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Answer 1 · 2018-05-13T09:33:34

# = 1 #

We know #1 - cos^2 theta = sin^2 theta #

#=> lim_( theta to 0 ) sin^2 theta / theta^2 #

#=> lim_(theta to 0 ) (sin theta / theta )^2 #

Use an approximation for #sin theta #

#=> sin theta = theta - theta^3 / (3! ) #

#=> lim_( theta to 0 )( ( theta - theta^3/(3! ) ) / theta ) ^2 #

#=> lim_ ( theta to 0 ) (1 - theta^2 / (3! ))^2 #

#= 1 #

Answer 2 · 2018-05-16T07:42:26

The limit is #=1#

We need

#lim_(theta->0)(1-cos^2theta)/theta^2=(1-1)/0=0/0#

which is indeterminate, so apply L'Hopital's rule

#lim_(theta->0)(1-cos^2theta)/theta^2=lim_(theta->0)((1-cos^2theta)')/((theta^2)')#

#=lim_(theta->0)(2costhetasintheta)/(2theta)#

#=lim_(theta->0)(sin2theta)/(2theta)#

#=lim_(theta->0)((sin2theta)')/((2theta)')#

#=lim_(theta->0)(2cos2theta)/(2)#

#=1#