Sunday, 8 March 2015

limits - How to evaluate limxto0fracsin2left(fracx2right)fracx24ex2+ex22?

limx0sin2(x2)x24ex2+ex22L=limx0sinx2cosx212x2xex2+(2x)ex2=limx0sinx2cosx212x2xex22xex2L=limx012cos2x212sin2x212(2x)(2x)ex2(2x)(2x)(ex2)=limx012cos2x212sin2x2124x2ex2+4x2ex2L=limx012(sinx2cosx2)12(sinx2cosx2)(4x2)(2x)ex2+(4x2)(2x)(ex2)=limx0sinx2cosx28x3ex28x3ex2



After evaluating the limit as x0, I noticed that the problem comes up to be in an indeterminate form of 0/0. I immediately utilized the L'Hospital Rule by differentiating both the numerator and denominator.



However, after using L'Hospital rule for 5-6 times, I noticed that the question will go through a loop of 0/0 indeterminants.



In my second attempt,
I have tried multiplying exp(x2) in both the numerator and denominator with hopes to balance out the exp(x2). However, an indeterminant is 0/0 still resulting.




Any help would be appreciated, thank you all!

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