Use the method of your choice to evaluate the following limits. lim(x,y)→(0,π/2)1−cos⁡xy4x2y3

Question

Use the method of your choice to evaluate the following limits.  lim(x,y)→(0,π/2)1−cos⁡xy4x2y3

nitruraviX · Accepted Answer

We have to solve the limit:  lim(x,y)→(0,π/2)1−cos⁡xy4x2y3  We know the formula,  cos2⁡x−sin2⁡x=cos⁡2x  1−cos⁡2x=2sin2⁡x  for numerator,  1−cos⁡xy=2sin2⁡(xy2)  Therefore,  lim(x,y)→(0,π2)1−cos⁡xy4x2y3=lim(x,y)→(0,π2)2sin2⁡(xy2)4x2y3  We know the formula  limx→0sin⁡xx=1  limx→0sin2⁡xx2=1  Multiplying and dividing by (xy2)2  lim(x,y)→(0,π2)2sin2⁡(xy2)4x2y3=lim(x,y)→(0,π2)2sin2⁡(xy2)(xy2)2×(xy2)24x2y3  =lim(x,y)→(0,π2)1×(xy2)24x2y3  =lim(x,y)→(0,π2)x2y244x2y3  =lim(x,y)→(0,π2)x2y24×4x2y3  =lim(x,y)→(0,π2)y216y3  =lim(x,y)→(0,π2)116y  =116×π2  =216π  =18π  Hence, value of limit is 18π

Jeffrey Jordon · Accepted Answer

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Use the method of your choice to evaluate the following limits. lim(x,y)→(0,π/2)1−cos⁡xy4x2y3

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