Calculus and AnalysisUse polar coordinates to find the limit.
2 months ago
Use polar coordinates to find the limit. [If are (r, theta) polar coordinates of the point (x,y) with r>=0, note that as (x,y) tends to (0,0).] lim (x,y) tends to (0,0)(x^(2)+y^(2))ln(x^(2)+y^(2))
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cyndilauperconcertadelaide Staff answered 2 months ago
Substitute into polar coordinates, that is: x=rcos 0 y=rsin 0 Therefore lim_((x,y) -> (0,0))(x^(2)+y^(2))ln(x^(2)+y^(2))=lim_(r -> 0^(+))(r^(2)cos^(2)0+r^(2)sin^(2)0)ln(r^(2)cos^(2)0+r^(2)sin^(2)0) =lim_(r -> 0^(+))r^(2)(cos^(2)0+sin^(2)0)ln(r^(2)(cos^(2)0+sin^(2)0)) =lim_(r -> 0^(+))r^(2)ln r^(2) =lim_(r -> 0^(+))(ln r^(2))/(r^(-2)) now we have (-infty)/(infty) type of limit, and we can use L'Hopital rule: =lim_(r -> 0^(+))((∂)/(∂ r)(ln r^(2)))/((∂)/(∂r)(r^(-2))) =lim_(r -> 0^(+))((1)/(r^(2))*2r)/(-2r^(-3)) =lim_(r -> 0^(+))(r^(4))/(-2r^(2)) =lim_(r -> 0^(+))(r^(2))/(-2) =0 Result: lim_((x,y) -> (0,0))(x^(2)+y^(2))ln (x^(2)+y^(2))=0
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