Evaluate limx→0sin(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
limx→0sin(x2)sin2(x)=limx→0(xsinx)2sin(x2)x2=1
No comments:
Post a Comment