Use polar coordinates to find the limit. [Hint: Let x = r cos and y = r sin , and note that (x, y) (0, 0) implies r 0.] lim_((x,y)->(0,0)) (x^2-y^2)/sqrt(x^2+y^2)

alesterp 2021-02-05 Answered
Use polar coordinates to find the limit. [Hint: Let x=rcosandy=rsin , and note that (x, y) (0, 0) implies r 0.] lim(x,y)(0,0)x2y2x2+y2
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Expert Answer

stuth1
Answered 2021-02-06 Author has 97 answers
Given the multivariable limit function:
lim(x,y)(0,0)x2y2x2+y2
For polar coordinate system:
x=rcosθ,y=rsinθ
lim(x,y)(0,0)x2y2x2+y2=limr0r2cos2θr2sin2θr2cos2θ+r2sin2θ
lim(x,y)(0,0)x2y2x2+y2=limr0r2(cos2θr2sin2θ)r2(cos2θ+r2sin2θ)
We know the trigonometry identities
(cos2θsin2θ)=cos2θ
(cos2θ1sin2θ)=1
lim(x,y)(0,0)x2y2x2+y2=limr0r2cos2θr
lim(x,y)(0,0)x2y2x2+y2=limr0rcos2θ
lim(x,y)(0,0)x2y2x2+y2=0
Hence the limit of the given function is zero.
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