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I have two questions regarding this subject. Hope you can help me.

Consider the limit of the function $f(x,y) = \frac{x^2y}{x^4+y^2}$ as (x,y) approaches (0,0): $$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^4 + y^2} = \lim_{r \to 0}\frac{r^2\cos^2\theta(r\sin\theta)}{r^4\cos^4\theta + r^2\sin^2\theta}$$ I tried the paths $y=x^2$ and $x=0$ and found different limits which means that the limit of the function doesn't exists. Now, I want to verify that through the polar coordinates. After the simplification, the polar equation reduces to: $$\lim_{r \to 0} \frac{r (\cos^2\theta\sin\theta)}{r^2\cos^4\theta + \sin^2\theta}$$ At first I thought that this limit is equal to $0$ for all lines except the lines $\theta = 0$ and $\theta= \pi$ because I thought these lines would make $sin \theta$ equal to $0$ and that would cause the indeterminate form $0/0$, but after I thought about it, I came to another conclusion:

Since this is a limit, $r$ only approaches to $0$ and it is never actually $0$, then the numerator of the limit is $0$ because of the $sin\theta$ part, and its denominator is not $0$ since $r$ is not $0$. Then, the limit should be just $0$. Is that right ?

My other question is about using different paths on polar coordinates. For example, consider the function $r = sin\theta$. If I use this path, the equation becomes: $$\lim_{r \to 0} \frac{sin^2\theta cos^2\theta}{sin^2\theta \cos^4\theta + \sin^2\theta}$$ Simplifying ${sin^2\theta}/{sin^2\theta}$, we get, $$\lim_{r \to 0} \frac{cos^2\theta}{cos^4\theta + 1}$$ My question is: Can I use the possible $\theta$ values in this expression? For example, Because $r=sin\theta$ and $r$ approaches to $0$, $\theta$ can either be $0$ or $\pi$. Since we only have the even powers of $cos\theta$, we can just assume that $cos^2\theta$ is $1$, which makes the limit equal to: $$\lim_{\theta \to 0} \frac{1}{1 + 1}= 1/2$$ Am I allowed to use $\theta$ value in the limit or is my work after the limit $\lim_{r \to 0} \frac{cos^2\theta}{cos^4\theta + 1}$ just wrong ?

Edit: I was told that my question is a possible duplicate of another question. I am very new at this website, so I'm not sure if this is the right way to explain why my question is not a duplicate but I'm gonna try to explain why. Mine and the other question is about the same limit; however, in the other question, the user asked why the limit doesn't exist, while I already know why, I just want to verify this in terms of polar coordinates. I also want to know if using the value of $\theta$ in different paths is valid, which is a topic the other question doesn't mention.

Berkay Döner
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  • In your first polar equation the $r$ in the numerator cancels and you are left with $r$ in the denominator. As $r$ gets smaller the entire limit gets arbitrarily large. – John Douma Aug 31 '18 at 19:46
  • I like your idea to say let $r = \sin \theta$ but since we are calculating the limit as $r$ goes to $0$ it seems cleaner to me to consider the path $\theta = r$ and then let $r$ go to $0.$ Regardless, on some paths to the origin the limit equal $\frac 12$ and on others it equals $1.$ And the limit only exists if it has the same value on all paths. – Doug M Aug 31 '18 at 20:05
  • You are doing right, Berkay. Just let $\theta$ go to $0$. In the last limit you may want to replace $r$ by $\theta.$ However $r=\sin \theta$ is just one path, and this does not prove that the limit exists. – Viera Čerňanová Aug 31 '18 at 20:17
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    Possible duplicate of Limit $\frac{x^2y}{x^4+y^2}$ is found using polar coordinates but it is not supposed to exist. – Jyrki Lahtonen Aug 31 '18 at 20:29
  • @DougM If I consider the path r=θ then the limit would become limθ→0θsinθcos2θθ2cos4θ+sin2θ and after letting cosθ=1 and applying L'Hopital's rule I found that the limit is 1/2. So, when considering r=sinθ or your suggestion r=θ or any path through the origin, we can regard $\theta$ as the inverse function of r=f(θ) and find its value by letting $r$ go 0, and use this value to find the limit, am I right? – Berkay Döner Aug 31 '18 at 22:31
  • @Maam I edited my question. I know different paths to origin gives different results and thats why the limit doesn't exists. I just want to ask you the same thing I asked to Doug M. When dealing with these limits, we can regard θ as the inverse function of r=f(θ) and find its value by letting r go 0, and use this value to find the limit, am I right? – Berkay Döner Aug 31 '18 at 22:37

1 Answers1

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Yes we have that for $\theta=0,\pi$

$$ \frac{r (\cos^2\theta\sin\theta)}{r^2\cos^4\theta + \sin^2\theta}=0$$

Yes along the path $r=\sin \theta$ we have

$$\lim_{r \to 0} \frac{\cos^2\theta}{\cos^4\theta + 1}= \frac12$$

since $\cos^2 \theta \to 1$, indeed the path $r=\sin \theta$ represents a circle centered at $(0,1/2)$ with radius $1/2$ indeed

$$r=\sin \theta \iff r=y/r \iff r^2-y=0 \iff x^2+y^2-y=0 $$$$\iff x^2+(y-1/2)^2=(1/2)^2$$

and at $(0,0)$ we have

$$y=\frac12-\frac12\sqrt{1-4x^2}$$

therefore fo $x \to 0$

$$y \approx \frac12-\frac12(1-2x^2)=x^2$$

and the parametrization behaves as $x=t$ $y=t^2$ with $t\to 0$ which indeed gives the same limit.

user
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  • Thank you! So, any straight line with a constant $\theta$ value gives the limit $0$. However, $r=sin\theta$ or your example the parabola $y=x^2$ gives the limit 1/2, which is different from $0$ and this causes the limit not to exist. Can you please correct me if I am wrong ? – Berkay Döner Aug 31 '18 at 22:52
  • @BerkayDöner Yes exactly, when th elimit exist it must be the same for any path thus if we find at least 2 paths with different limits then the limit doesn't exist. The fact is used indeed to prove when limit do not exist. – user Sep 01 '18 at 00:42
  • @BerkayDöner I've added some detail more for the parametrization in the case $r=\sin \theta$. – user Sep 01 '18 at 07:41
  • Thank you, now it is clearer to me. I just want to ask you a final question since other users didn't answer it. I want to generalize my question. So, to calculate any multivariable limit where $(x,y)$ goes to $(0,0)$, we can convert it to polar coordinates and consider a path, which was $r=sin\theta$ in my question, and regard $\theta$ as the inverse function of $r=f(\theta)$. Then, we can find the limit of this function when $r$ goes to $0$ and use this value to find the value of the limit. Like I did in my question by using the $\theta=0$ and $\theta=\pi$ values. Is that right ? – Berkay Döner Sep 01 '18 at 08:36
  • Though, I know we need to find the same value for any path to limit to exist. – Berkay Döner Sep 01 '18 at 08:39
  • @BerkayDöner Pay attention to that when dealing with limits of several variables: to show that the limit exist we can't show that by considering all the paths (there are infinite paths indeed) but we need some other method (inequality and squeeze, Taylor, etc.); to show that the limit doesn't exist the strategy is almost always to find two or more paths with different limit. – user Sep 01 '18 at 08:43
  • Yeah, I know we have to consider two or more paths with different limits to show that the limit doesn't exist. I just wanted to say, we can find these limits by following a path and using the appropriate values of $\theta$ in these paths. English is not my first language so I sometimes have a hard time expressing myself. – Berkay Döner Sep 01 '18 at 09:44