limx→0sin2(x2)−x24ex2+e−x2−2L=limx→0sinx2cosx2−12x2xex2+(−2x)e−x2=limx→0sinx2cosx2−12x2xex2−2xe−x2L=limx→012cos2x2−12sin2x2−12(2x)(2x)ex2−(2x)(−2x)(e−x2)=limx→012cos2x2−12sin2x2−124x2ex2+4x2e−x2L=limx→012(−sinx2cosx2)−12(sinx2cosx2)(4x2)(2x)ex2+(4x2)(−2x)(e−x2)=limx→0−sinx2cosx28x3ex2−8x3e−x2
After evaluating the limit as x→0, 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(x−2). However, an indeterminant is 0/0 still resulting.
Any help would be appreciated, thank you all!
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