Use the method of your choice to evaluate the following limits. lim_{(x,y)rightarrow(0,pi/2)}frac{1-cos xy}{4x^2y^3}

Tyra 2020-11-27 Answered
Use the method of your choice to evaluate the following limits.
lim(x,y)(0,π/2)1cosxy4x2y3
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Expert Answer

nitruraviX
Answered 2020-11-28 Author has 101 answers
We have to solve the limit:
lim(x,y)(0,π/2)1cosxy4x2y3
We know the formula,
cos2xsin2x=cos2x
1cos2x=2sin2x
for numerator,
1cosxy=2sin2(xy2)
Therefore,
lim(x,y)(0,π2)1cosxy4x2y3=lim(x,y)(0,π2)2sin2(xy2)4x2y3
We know the formula
limx0sinxx=1
limx0sin2xx2=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
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Jeffrey Jordon
Answered 2022-04-01 Author has 2087 answers

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