Find the solution of limit lim(x,y)→0,0x2+y2x2+y2 by using the polar coordinates system.

lim(x,y)→0,0(x2+y2(x2+y2))=lim(x,y)→0,0f(x,y), where f(x,y)=(x2+y2).Using polar form x=rcos⁡0,y=rsin⁡0 we have f(r,0)=rr2(cos20+sin20)=rr2 Now limr→0,0→0f(r,0)=limr→0rr2=∞ Therefore lim(x,y)→0,0x2+y2x2+y2=∞

Question: "Find the solution of limit lim(x,y)→0,0x2+y2x2+y2 by using..."

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necessaryh

Answered question

2021-01-17

Find the solution of limit $lim_{(x, y) \to 0, 0} \frac{\sqrt{x^{2} + y^{2}}}{x^{2} + y^{2}}$ by using the polar coordinates system.

Answer & Explanation

un4t5o4v

Skilled2021-01-18Added 105 answers

lim_{(x, y) \to 0, 0} (\sqrt{x^{2} + y^{2}} (x^{2} + y^{2})) = lim_{(x, y) \to 0, 0} f (x, y)

,
where

f (x, y) = (\sqrt{x^{2} + y^{2}}) . Using polar form x = r \cos 0, y = r \sin 0

we have

f (r, 0) = \frac{r}{r^{2} (\cos^{2} 0 + \sin^{2} 0)} = \frac{r}{r^{2}}

Now

lim_{r \to 0, 0 \to 0} f (r, 0) = lim_{r \to 0} \frac{r}{r^{2}} = \infty

Therefore

lim_{(x, y) \to 0, 0} \frac{{\sqrt{x}}^{2} + y^{2}}{x^{2} + y^{2}} = \infty

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