Calculate the following limit:
limx→0(tanxx)1sin2x
I know the result must be 3√e but I don't know how to get it. I've tried rewriting the limit as follows:
limx→0eln(tanxx)1sin2x=limx→0e1sin2xln(tanxx)
From this point, I applied l'Hospital's rule but got 1 instead of 3√e.
Thank you!
Answer
limx→0logtanxxsin2xl′H=limx→0xtanxxsec2x−tanxx22sinxcosx=limx→01sinxcosx−1x2sinxcosx=
=limx→0x−sinxcosx2xsin2xcos2x⏟=x2sin22xl′H=limx→02sin2x⏞1−cos2x+sin2x12sin22x+xsin2xcos2x⏟=xsin4xl′H=limx→02sin2x2sin4x+4xcos4x=
l′H=limx→04cos2x12cos4x−16xsin4x=412=13
and the limit is e1/3
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