Tuesday, 9 July 2019

exponentiation - Calculate exponential limit involving trigonometric functions



Calculate the following limit:
limx0(tanxx)1sin2x



I know the result must be 3e but I don't know how to get it. I've tried rewriting the limit as follows:




limx0eln(tanxx)1sin2x=limx0e1sin2xln(tanxx)



From this point, I applied l'Hospital's rule but got 1 instead of 3e.



Thank you!


Answer



limx0logtanxxsin2xlH=limx0xtanxxsec2xtanxx22sinxcosx=limx01sinxcosx1x2sinxcosx=



=limx0xsinxcosx2xsin2xcos2x=x2sin22xlH=limx02sin2x1cos2x+sin2x12sin22x+xsin2xcos2x=xsin4xlH=limx02sin2x2sin4x+4xcos4x=




lH=limx04cos2x12cos4x16xsin4x=412=13



and the limit is e1/3


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