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Compute $\displaystyle\lim_{ (x,y)\to (0,0)}\dfrac{x^ny^m}{x^2+y^2}$

Determine with the conditions on $n$ and $m$ for which this limit exists and conditions for which this limit does not exist.

I found that when $x$ approaches $0$ the limit approaches $0$ and as $y$ approaches $0$ the limit also approaches $0$. As $x$ and $y$ approaches $x$ it seems like there are no conditions for $m$ and $n$ which will make the limit exist.

StubbornAtom
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Aubrey
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2 Answers2

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Hint: To get some insight into this expression, consider polar coordinates with $x = r\cos\theta$, $y = r\sin\theta$. Then

$$\frac{x^ny^m}{x^2 + y^2} = \frac{r^{m+n}\cos^n\theta\sin^m\theta}{r^2(\cos^2\theta + \sin^2\theta)} = \ ...$$

Simon S
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  • So when n and m are equal to 1 the limit exists but when they are not you are left with an r in the denominator which makes the limit not exist. Is that correct? – Aubrey Dec 04 '14 at 16:59
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    If $m + n < 2$ the limit is in trouble. But you should reconsider the cases $m + n = 2$ and $m + n > 2$. – Simon S Dec 04 '14 at 17:05
  • Okay so when m+n is between 0 and 2 (since both parameters must be equal) the limit does not exist because there is an r left in the denominator. Then m+n=2 the limit exists because the r's cancel out. When m+n is greater than 2 the limit also exists because there is an r left in the numerator. – Aubrey Dec 04 '14 at 17:07
  • If $m + n = 2$ is the limit independent of $\theta$? Is that a problem? – Simon S Dec 04 '14 at 17:10
  • When m+n=2 the function is bounded which means it is well defined. Does knowing that information help me at all? – Aubrey Dec 04 '14 at 17:13
  • If $\theta = 0$, then the expression is $0$ and the limit approaching along the line $\theta = 0$, i.e., the $x$-axis is zero. But if $\theta = \pi/4$, then the expression is $1/\sqrt{2}^{m+n}$ and the limit approaching $(0,0)$ from this direction is not zero. Hence in the case $m + n = 2$, the limit does not exist. – Simon S Dec 04 '14 at 17:16
  • Okay so since the limit changes when theta changes the limit does not exits. In the case that m+n>2 the limit would exist because there is an r value that will make the limit defined. Is that right? – Aubrey Dec 04 '14 at 17:19
  • You should convince yourself that that is true. – Simon S Dec 04 '14 at 17:21
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Try using polar coordinates ($x=r \cos(\theta), y=r\sin(\theta)$, then make $r$ tend to 0). Try also studying what happens along a straight line (do the substitution $x=ay$)

Octania
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