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Show that the following limit exists or does not exist (general example)

$$\lim \limits_{(x,y) \to (0,0)} \dfrac{e^{-x^2-y^2}-1}{x^2+y^2}$$


i) Direct substitution of $x=0$ , $y=0$ leads to indeterminate form of $\frac{0}{0}$

ii) Taking the limit along $x$ , $y$ axes and $y=x$ all result with the value $0$

iii) Convert to polar:

$$\lim \limits_{r \to 0^+} \dfrac{e^{-r^2}-1}{r^2}->\frac{0}{0}$$

$L'Hopital's$ $rule$

$$\lim \limits_{r \to 0^+} \dfrac{-2re^{-r^2}}{2r}=-1$$

So the limit exists and its value is -1


My questions:

  1. After converting the limit expression to polar, why is $\lim \limits_{r \to 0^+}$ instead of $\lim \limits_{r \to 0}$ ? Both have the same computation
  1. From the example above, how would I know if the limit $DNE$ when taking the limit after converting to polar? Would taking the limit of the polar converted expression $DNE$ or not give a finite number to know that the original limit $DNE$? This is of course if I chose to convert to polar without knowing that a different path gave a different limit.

  2. When would it be appropriate to covert to polar to show the existence of a limit when not told that it existed or not in the first place? Does converting to polar always work?


$$\lim \limits_{(x,y) \to (0,0)} \dfrac{{xy^4}}{x^2+y^8}$$

  • this limit $DNE$ as it has different limits along different paths namely $y=0$ and $x = y^4$, respectively 0 $≠$ $\frac{1}{2}$

Polar conversion: (this limit DNE, but polar conversion results in 0, a finite number)- to check

$$\lim \limits_{r \to 0^+} \dfrac{{rcosθ*r^4sin^4θ}}{r^2cos^2θ+r^8sin^8θ}$$

$$\lim \limits_{r \to 0^+} \dfrac{r^5({cosθ*sin^4θ})}{r^2(cos^2θ+r^6sin^8θ)}$$

$$\lim \limits_{r \to 0^+} \dfrac{r^3({cosθ*sin^4θ})}{cos^2θ+r^6sin^8θ}$$

$$\frac{0}{(cos^2(θ))}=0$$

The limit is 0

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  • $\begingroup$ Does converting to polar only work when (x,y) is approaching (0,0) or also work for say (x,y) is approaching a point like (-1,7)? $\endgroup$ – user314 Jun 23 '20 at 11:11
  • $\begingroup$ For the first question, I think it is just emphasising that $r$ cannot be negative so if you approach to $0$ you can only approach from the right $\endgroup$ – JustWandering Jun 23 '20 at 11:16
  • $\begingroup$ So does it mean that in polar coordinates, since it goes in counterclock-wise direction, it is same as $r->0^{+}$? $\endgroup$ – user314 Jun 23 '20 at 11:18
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For question 1, we take the limit as $r \to 0^{+}$ because in polar coordinates, $r$ represents the distance from the origin to point $(x, y)$ which is always non-negative.

For questions 2 and 3, keep in mind that we have

$$\lim_{(x, y) \to (0, 0)} \frac{e^{-x^2-y^2} - 1}{x^2 + y^2} = c$$

for some finite number $c$ if and only if

$$\lim_{r \to 0^{+}} \frac{e^{-r^2} - 1}{r^2} = c$$

In other words, the first limit is DNE if and only if the second one is DNE. Thus, if you manage to find some finite result $c$ for the second one, then you have also solved the first one. Sometimes, it is easier to evaluate limits in polar coordinates that in Cartesian coordinates so we take advantage of this when this applies.

An important note

Taking the limit along x , y axes and y=x all result with the value 0

It is important to note that in order for limit of a sequence to exist in a metric space like $\mathbb{R}^2$, all of its sub-sequences must also converge to that limit. That means that no matter how you walk your way to the limit, you must always arrive at the limit.

Hence, taking the limit along the $x$-axis, $y$-axis and the line $y = x$ is just one way to warn yourself early when the limit actually does not exist when these limits give different values.

But, if these limits all agree, this is not sufficient to say that the limit does converge to some finite number $c$ because there can be some distorted path to approach $(0, 0)$ for which a different limit can be computed.

However, the polar form takes into consideration all possible ways to walk to the origin because no matter how you approach $(0, 0)$, the distance from your point to $(0, 0)$ always converges to $0$, hence we have $r \to 0^{+}$.

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  • $\begingroup$ Also if the original limit had (x,y) approaching some other point like (-2,9), would converting to polar not work then since r is not approaching the origin? $\endgroup$ – user314 Jun 23 '20 at 11:23
  • $\begingroup$ For this specific limit in this question, yes, you cannot take advantage of the nice polar form to evaluate your limit easily if $(x, y)$ does not approach the origin. $\endgroup$ – Royce Pacibe Jun 23 '20 at 11:30
  • $\begingroup$ Wait I put another example in my question now, and the limit does not exist by taking limit of different paths, but when converting to polar, the limit is 0, a finite number $\endgroup$ – user314 Jun 23 '20 at 11:30
  • $\begingroup$ Unfortunately, for that supposed counter-example, you can't take advantage of that nice polar form. Usually, the hint is the presence of $x^2 + y^2$. This is because $r$ is equal to the distance of $(x, y)$ to the origin, i.e. $r = \sqrt{x^2 + y^2}$ which is equivalent to $r^2 = x^2 + y^2$, the classic Pythagorean theorem. :) $\endgroup$ – Royce Pacibe Jun 23 '20 at 11:33
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    $\begingroup$ Actually, there is a problem with the evaluation of your limit. The end result is $0/\cos^{2}\theta$ which is not always zero for all angles $\theta$. For instance, if you approach along the angle $\theta = \pi/2$, the expression evaluates to the indeterminate form $0/0$. However, in the first question, you had that nice $x^2 + y^2$ in it that you eliminated the $\theta$'s entirely when converted to polar form, so we were able to evaluate it easily to $-1$ without problems. :) $\endgroup$ – Royce Pacibe Jun 23 '20 at 11:44
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If you restrict the polar argument to the range $[0,2\pi)$, the Cartesian-to-polar transformation is a bijection. Hence whatever limit computation you perform in polar coordinates gives exactly the same conclusion as when computed in Cartesian.

Polar coordinates are used for convenience when a polar symmetry (like in your example) or a significant simplification is apparent.

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