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Find the following limit:$$\lim_{(x,y) \to (0,0)} \dfrac{x^ay^b}{x^2+y^2} a,b \ge 0$$ I tried the $x=0$ path for which the limit is $0$. Then I took $x=y$ for which we have three cases: $$\lim_{x \to 0} \dfrac{x^{a+b}}{2x^2}=\lim_{x \to 0}\dfrac{1}{2}x^{a+b-2}$$ For $a+b-2<0$ The above limit is infinity. So the multivariable limit does not exist. For $a+b-2>0$ The above limit is 0. Should I take other path in this case? If so, how? For $a+b-2=0$ I am not sure if I calculated the limit right. Is it $\dfrac{1}{2}$? Becuse you get $x^0 = 1$

Tell me if my approach is wrong. I want to understand well this stuff!

Another approach: Using polar coordinates: $x=r\cos\theta, y=r\sin\theta$

$$\dfrac{x^ay^b}{x^2+y^2}= r^{a+b-2}(\cos\theta)^a(\sin\theta)^b$$So if $a+b>2$, then when $r\to0$ the limit is $0$ for all $\theta$, since $|(\cos\theta)^a(\sin\theta)^b|<1$. If $a+b=2$ then you have $(\cos\theta)^a(\sin\theta)^b$ which depends on $\theta$. Should I just plug in numbers for $\theta$ and see that I get different values, therefore the limit does not exist? What about the case when $a+b<2$. Gives me infinity except when one of the two trigonometric functions are $0$.

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    Just to make sure, do $a,b $ have to be integers? Or are they any real positive numbers? – 2017-01-10
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    Any real positive numbers – 2017-01-10

2 Answers 2

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This is how I would go about solving it, continuing what you did, given that I didn't think of polar coordinates:

  • for $a+b > 2$ you didn't prove it exists, only that it should be 0 if it exists. This is how I'd prove it: (assuming absolute values of these quantities and omitting the limit notation)

$$x \leq ||(x, y)||; y \leq ||(x, y)|| \implies \frac {x^ay^b}{x^2 + y^2} \leq \frac {||(x, y)||^{a+b}}{||(x, y)||^2} = ||(x, y)||^{a+b-2} \to 0$$

Because $a+b-2 > 0$, where $||(x, y)||$ is the usual norm of the vector $(x, y) $, given by $\sqrt {x^2 + y^2} $. Therefore the limit goes to 0.

  • For $a+b \leq 2$ I'd set $y = mx $ to get: (again omitting the limit part)

$$\frac {x^ay^b}{x^2 + y^2} = \frac {x^a(mx)^b}{x^2 + (mx)^2} = \frac {m^bx^{a+b}}{(m^2+1)x^2} = \frac {m^b}{(m^2+1)x^\epsilon}, \epsilon \geq 0$$

If $\epsilon = 0$, the limit does not exist because it changes with the slope $m $ of the line you use to approach the origin:

$$\epsilon = 0 \implies \frac {m^b}{(m^2+1)x^\epsilon} = \frac {m^b}{(m^2+1)}$$

If $\epsilon > 0$ then the limit is infinite as you get something of the form

$$\lim_{t \to 0} \frac {k}{t} $$

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    I can infer from your solution that by $||(x,y)||$ you mean the maximum value between $x$ and $y$. Is that correct? – 2017-01-10
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    Actually no, $(x, y)$ is a vector with coordinates $x $ and $y $ and $||(x, y)||$ is the norm ("length") of the vector, given by the usual expression $\sqrt{x^2 + y^2} $ – 2017-01-10
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Hint: write $x=rcos(u), y=rsin(u)$ so the limit is $lim_{r\rightarrow 0}r^{a+b-2}cos(u)^asin(u)^b$.