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))

1 Answers

Best Answer

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