Use polar coordinates to find the limit. [if (r, θ) are polar coordinates of the point (x, y) with r ⥠0, note that r â 0+ as (x, y) â (0, 0).] (if an answer does not exist, enter dne.) lim (x, y)â(0, 0) 7eâx2 â y2 â 7 x2 + y2

Respuesta :

I'll take a wild guess and suggest the limit is supposed to be

[tex]\displaystyle\lim_{(x,y)\to(0,0)}\frac{7e^{-x^2-y^2}-7}{x^2+y^2}[/tex]

Converting to polar coordinates, we take [tex]x^2+y^2=r^2[/tex] with [tex]x=r\cos t[/tex] and [tex]y=r\sin t[/tex]. This yields

[tex]\displaystyle\lim_{(r,t)\to(0,0)}\frac{7e^{-r^2}-7}{r^2}[/tex]

Most important is that the limand is independent of [tex]t[/tex]. Evaluating directly yields [tex]\dfrac00[/tex], so we apply L'Hopital's rule:

[tex]\displaystyle\lim_{r\to0}\frac{7e^{-r^2}-7}{r^2}=\lim_{r\to0}\frac{-14re^{-r^2}}{2r}=-7\lim_{r\to0}e^{-r^2}=-7[/tex]
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