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I was trying to evaluate the following limit:

$$ \lim_{(x,y) \to (0,0)} \frac{x^2 y e^y}{x^4 + 4y^2}. $$

I attempted to convert this into polar coordinates. Substituting $x = r \cos \theta$ and $y = r \sin \theta$, the expression becomes:

$$ \frac{(r^2 \cos^2 \theta)(r \sin \theta) e^{r \sin \theta}}{(r^4 \cos^4 \theta) + 4(r^2 \sin^2 \theta)}. $$

When $ \theta = 0 $, the function itself is $ 0 $, and when $ \theta \neq 0 $, the limit appears to approach $ 0 $ as $ r \to 0 $. Thus, I concluded that the limit converges to $ 0 $ for all values of $ \theta $.

However, when I approached the limit along the path $ y = mx^2 $, I found that the limit does not exist.

Could someone point out the error in my reasoning? Some suggested that I violated the property of limits because I directly substituted $ r = 0 $ into the expression:

$$ \frac{0}{0 + 4 \sin^2 \theta}. $$

They claimed this approach is not rigorous, but I do not understand why. Is there a problem with how I handled the substitution in polar coordinates?

LACKHOLE
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  • You concluded that “for each $\theta$ the limit as $r\to 0^+$ is $0$”. This is correct, but this says NOTHING about whether the limit $(x,y)\to (0,0)$ exists (but it does tell you that if the limit exists, then it is $ 0$). See my answer here; also another way of expressing my uniformity condition (3) from the link is in this answer (though a mild edit should be made to specify that $g$ is a real-valued function only defined on some interval $(0,\delta)$, rather than all of $\Bbb{R}$). – peek-a-boo Dec 16 '24 at 00:58
  • Also, your case distinction is wrong. You should separate out the case when $\sin\theta=0$ vs when $\sin\theta\neq 0$. In the former case the whole function is $0$, while in the latter, we have essentially $r^3/r^2=r$ which goes to $0$. – peek-a-boo Dec 16 '24 at 01:06
  • Then, how can we determine whether analyzing the limit of a function in polar coordinates along straight-line paths with

    θ fixed is useful or not?

     – LACKHOLE Dec 16 '24 at 01:18
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    If limits along different straight lines is different, then the limit doesn’t exist. But if all straight line limits exist and are equal, that’s still insufficient to conclude anything. Notice that checking along straight lines is something you can do even without polar coordinates, which is why I keep emphasizing that polar coordinates are an unnecessary distraction. – peek-a-boo Dec 16 '24 at 01:49
  • How did you get $4 \sin^2 \theta$ in the last formula in the question? If you set $r=0$ then $4(r^2 \sin^2 \theta)$ should evaluate to $0,$ shouldn't it? You would then get $0/0$ as your result, from which you should not conclude anything. – David K Dec 18 '24 at 01:20
  • Another way to ask the question is: when are polar coordinates useful in a multivariable limit? There are answers to that question here and here. – David K Dec 18 '24 at 01:33
  • "Thus, I concluded that the limit converges to $ 0 $ for all values of $ \theta $." And that's why this particular attempt to use polar coordinates is no good for establishing a limit. You only looked at paths along straight lines to the limit point. Just don't do that. – David K Dec 18 '24 at 01:41

2 Answers2

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The $e^y$ factor is irrelevant (since it is approximately $1$). Let $u=x^2$ and $v=2y$. Then you are looking at the familiar classic $$g(u,v)=\frac12\cdot\frac{uv}{u^2+v^2},$$ which varies between $-1/4$ and $1/4$ as $(u,v)$ tends to $(0^+,0)$.

Ted Shifrin
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You can use modified polar coordinates, namely $$x^2=r\sin \theta,\quad 2y=r\cos\theta$$ Then $0\le \theta\le \pi$ and we get $$ {r\sin\theta \,{r\over 2}\cos\theta\,e^{(r\cos\theta)/2}\over r^2}={1\over 2}e^{(r\cos \theta)/2}\sin\theta\cos\theta$$ The limit, when $r\to 0^+,$ does not exist.

Ryszard Szwarc
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