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Use polar coordinates to find the limit. [if (r, ?) are polar coordinates of the point (x, y) with r ? 0, note that r ? 0+ as (x, y) ? (0, 0).] (if an answer does not exist, enter dne.) lim (x, y)?(0, 0) (x2 + y2) ln(x2 + y2)

Respuesta :

LammettHash
[tex]\displaystyle\lim_{(x,y)\to(0,0)}(x^2+y^2)\ln(x^2+y^2)=\lim_{(r,\theta)\to(0^+,0)}r^2\ln(r^2)=\lim_{r\to0^+}r^2\ln(r^2)[/tex]

since the limand is independent of [tex]\theta[/tex].

Taking [tex]\rho=\dfrac1r[/tex], we have (by L'Hopital's rule in the first step below)

[tex]\displaystyle-\lim_{\rho\to\infty}\frac{\ln\rho^2}{\rho^2}=-\lim_{\rho\to\infty}\frac{\frac{2\rho}{\rho^2}}{2\rho}=-\lim_{\rho\to\infty}\frac1{\rho^2}=0[/tex]

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