Problem

limx04sinxsin2x2x3

Answer

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Answer

Apply L'Hopital's rule and find the limit as x approaches 0: limx04sin(x2)2cos(x)+4sin(x2)sin(x)cos(x2)3x2=1

Steps

Step 1 :Find the derivatives of the numerator and the denominator: ddx(4sinxsin2(x2))=4sin(x2)2cos(x)+4sin(x2)sin(x)cos(x2) and ddx(x3)=3x2

Step 2 :Apply L'Hopital's rule and find the limit as x approaches 0: limx04sin(x2)2cos(x)+4sin(x2)sin(x)cos(x2)3x2=1

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