Sunday, 21 April 2019

calculus - Finding limxto0fracsin(x2)sin2(x) with Taylor series




Evaluate limx0sin(x2)sin2(x).



Using L'Hospital twice, I found this limit to be 1. However, since the Taylor series expansions of sin(x2) and sin2(x) tell us that both of these approach 0 like x2, I'm wondering if we can argue that the limit must be 1 via the Taylor series formal way?


Answer



limx0sin(x2)sin2(x)=limx0(xsinx)2sin(x2)x2=1


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