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I don't know how to approach this exercise, I tried looking at directional paths $y=mx$ and found out that if $\alpha+\beta-2=0$ then the limit doesn't exist.

From the same directional paths I think that the condition $\alpha+\beta-2<0$ means the limit doesn't exist since if I approach $0$ from the right then the limit must be $+\infty$ and from the left is $-\infty$. Though I really don't know if I used the definition correctly, is this part right?

I'm lost at $\alpha+\beta-2>0$, I know the limit must be $0$ but I don't know how to get extra conditions so that I get a for sure convergent limit.

rtybase
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Sebastian Cor
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1 Answers1

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Often in these sorts of questions it helps to transform to polar coordinates so the limit is then of just $r \to 0$. Also, if there is anything left depending on $\theta$ (but be careful to make sure that they don't cancel out in some more complicated situations, where you may want to try a few values of $\theta$ to confirm you get different values), it usually means the limit doesn't exist as different paths will give different results.

Thus, let $x = r\cos \theta$ and $y = r\sin \theta$, and noting that $x^2 + y^2 = r^2$, the limit becomes

\begin{align} \lim_{r \to 0} \frac{\left(r\cos \theta\right)^{\alpha} \left(r\sin \theta\right)^{\beta}}{r^2 + \left(r\cos \theta\right)\left(r\sin \theta\right)} & = \lim_{r \to 0} \frac{r^{\alpha + \beta}\left(\cos \theta\right)^{\alpha} \left(\sin \theta\right)^{\beta}}{r^2\left(1 + \left(\cos \theta\right)\left(\sin \theta\right)\right)} \\ & = \lim_{r \to 0} r^{\alpha + \beta - 2} \frac{\left(\cos \theta\right)^{\alpha} \left(\sin \theta\right)^{\beta}}{1 + \left(\cos \theta\right)\left(\sin \theta\right)} \tag{1}\label{eq1} \end{align}

You now have $3$ basic cases to consider depending on whether $\alpha + \beta - 2$ is $\gt 0$, $= 0$ or $\lt 0$. I believe you should be able to finish the rest of it now. However, be careful with the $2$nd & $3$rd cases in terms of how the values may change depending on the path taken to the origin, as you've already determined in your question text, and with what you've written also indicates you've already got most of this figured out.

Update: As discussed in the comments below, an additional required condition fora limit to exist is that $\alpha$ and $\beta$ must both be non-negative. Also, as I later realized, as we are dealing with real powers & potentially negative values of $x$ and $y$, along with the implicit assumption that the expression in the limit is restricted to resulting in only real values, then there's the issue of for which values may there be a limit among the non-integral values of the powers. This subject is discussed in various MSE questions such as Non-integer powers of negative numbers and Dealing with non-integral powers on negative numbers, plus on StackExchange such as Calculating pow() of negative number to non-integral exponent. In general, I believe you will need to restrict the powers to be rational values where, in lowest terms, the denominator is an odd integer (e.g., $\frac{1}{3}$ and $\frac{2}{3}$ are allowed, as the cube root of a negative is defined as a negative, but $\frac{1}{2}$ is not allowed as the square root of a negative is not real), with this including integers as the denominator would be $1$. This rational number restriction is discussed further in Quora at What happens when a negative number is raised to a non-integer power?.

John Omielan
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  • how do I deal with the denominator $1+(\cos\theta)(\sin\theta)$ in the $\alpha + \beta -2>0$ case, since $(\cos\theta)(\sin\theta)=\frac{\sin2\theta}{2}$ and $\sin2\theta=-1$ for $\theta=\frac{3}{4\pi}$, normally the idea would be to see if I can bound the expression multiplying $r^{\alpha+\beta-2}$ term but I don't think you can because of that, right? – Sebastian Cor Feb 22 '19 at 01:52
  • Also in the case for $\alpha+\beta-2=0$ I don't think you need to be that carefull because the limits is equal to an expression dependant on $\theta$ so it clearly can't exist. – Sebastian Cor Feb 22 '19 at 01:54
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    @SebastianCor In the $\alpha + \beta - 2 \gt 0$ case, the power of $r$ is positive, so as $r \to 0$, it goes to $0$. Since as you say that $\left(\cos \theta\right)\left(\sin \theta\right) = \frac{\sin 2\theta}{2}$, so the denominator is bound between $1 - \frac{1}{2} = \frac{1}{2}$ and $1 + \frac{1}{2} = \frac{3}{2}$, and the numerator is also bounded as $-1 \le \cos \theta, \sin \theta \le 1$, so the entire fraction is bounded. Thus, the entire expression will go to $0$. Does this make sense? Also, for the case $\alpha + \beta - 2 = 0$, you are correct that the limit doesn't exist. – John Omielan Feb 22 '19 at 01:59
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    @SebastianCor One other thing to be careful of is to consider what happens if $\alpha$ or $\beta$ is $\lt 0$, as you would then get values like $\frac{1}{x}$ and $\frac{1}{y}$, which can possibly affect the results even when $\alpha + \beta - 2 \gt 0$. – John Omielan Feb 22 '19 at 02:04
  • Oh you're right it is bounded I totally missed that. – Sebastian Cor Feb 22 '19 at 02:08
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    @SebastianCor Note, however, that the fraction is not necessarily bounded in the numerator if $\alpha$ or $\beta$ is $\lt 0$. I forgot to mention in my earlier comment that it was implicitly assume that $\alpha$ and $\beta$ are both $\ge 0$. This is why I wrote my next comment above. – John Omielan Feb 22 '19 at 02:11
  • So in the cases where either one of $\alpha$ or $\beta$ <0 is the test inconclusive ? – Sebastian Cor Feb 22 '19 at 02:17
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    If $\alpha \lt 0$, then going along certain paths, such as where $y = x^{-\frac{\alpha}{2\beta}}$, for example, causes a problem as the numerator becomes $x^{\frac{\alpha}{2}}$, which goes to $\infty$ as $x \to 0$. Similarly having $\beta \lt 0$ also causes the same sort of problem. In these cases, you can't say the limit exists. – John Omielan Feb 22 '19 at 02:26
  • @SebastianCor As I state in my updated answer, you should also consider restrictions on what non-integral values of $\alpha$ and $\beta$ are allowed. – John Omielan Feb 22 '19 at 05:37
  • What do you mean by this? Isn't the condition $\alpha+\beta-2$ sufficiently precise? It doesnt look like the function has funny bussiness in non integer values such as $\alpha\in(0,1)$ – Sebastian Cor Feb 22 '19 at 10:50
  • @SebastianCor I'm not quite sure what the issue is. Your question title starts with "Let $\alpha,\beta \in \mathbb{R}$ ...". As such, I assumed that you would also need to consider how non-integral values of both $\alpha$ and $\beta$ would behave. As stated, you will run into problems with values such as $\alpha = 0.5$ because there are no real square roots of negative values of $x$, such as $x = -1$ giving $x^{\alpha} = \sqrt{-1}$. I was just trying to help ensure my answer to you was comprehensive & fully accurate. – John Omielan Feb 22 '19 at 15:45
  • @SebastianCor In my experience, unless stated otherwise, it's assumed the function involved in the limit is restricted to being real values. However, if complex values are also allowed, then you don't need to worry about $\alpha$ and $\beta$ possibly being non-integral values. I'm sorry I didn't state my implicit assumption originally. – John Omielan Feb 22 '19 at 18:26
  • Oh yea that makes sense, never thought about it! I'll just think of $x^\alpha$ defined on its maximum domain, thanks for the input. – Sebastian Cor Feb 23 '19 at 17:52