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For $f(x,y) = \frac{xy-1}{x^2 y^2-1}$, what is the limit as $(x,y)$ goes to $(1,1)$?

Since the denominator can be factored into $(xy-1)(xy+1)$ and then the $(xy-1)$'s in both the numerator and denominator can be cancelled, we are left with $f(x,y) = \frac{1}{xy+1}$ and substituting $x=1$ and $y=1$, the limit becomes $\frac{1}{2}$. Is this solution right? We had this question on our test todayin Multivariable Caclulus, and some people put $1/2$ while others put $DNE$. Who is right?

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    $\begingroup$ The limit is 1/2. $\endgroup$ – Mike Earnest Sep 5 '18 at 15:55
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    $\begingroup$ Your answer and logic are correct; just note however that $f$ fails to exist along the parabolas $y = \pm 1/x$. $\endgroup$ – user296602 Sep 5 '18 at 15:58
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    $\begingroup$ @T.Bongers —when $y=+ 1/x$, then don't we have $f(x,y) = 1/2$? $\endgroup$ – space Sep 5 '18 at 16:01
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    $\begingroup$ The question is a bit unclear. Normally when discussing limits of a function at some point $P$ we require that the function is defined in some punctured heighborhood of $P$. This condition is not met here, and therefore we are justified to conclude that the limit DNE. However, some authors may (conceivably) define the limit of a function $f:A\to\Bbb{R}$ at a point only to involve neoghborhoods like $(B(P,r)\cap A)\setminus\{P\}$. If you are using such a definition, then your answer is correct, and the limit is $1/2$. We cannot be sure, but my knee-jerk reaction is to claim DNE. $\endgroup$ – Jyrki Lahtonen Sep 5 '18 at 18:12
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    $\begingroup$ @JyrkiLahtonen In baby Rudin, in order to discuss $\lim_{x\to p}f(x)$ of a function with domain $E$, it is only required to have $p$ be a limit point of $E$. $\endgroup$ – Mike Earnest Sep 6 '18 at 20:42
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Yes it is right, indeed we have that as $(x,y) \to (1,1)$ also $t=xy \to 1$ therefore we reduce to a limit for a single variable that is

$$\frac{xy-1}{x^2 y^2-1}=\frac{t-1}{t^2-1}=\frac{t-1}{(t+1)(t-1)}=\frac{1}{t+1}\to \frac12$$

As noticed in the comments $f(x,y)$ fails to exist along the hyperbolas $y=\pm\frac1x$ but it doesn't matter since these points are not in the domain.

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  • $\begingroup$ Every single neighborhood of $(1,1)$ contains points on the hyperbola $y=1/x$. At least some sources (all of those I have used recently) require in the definition of a limit of a function at a point $P$ that the domain should contain a punctured neighborhood of $P$. Of course, it also makes perfect sense to only use the subspace topology of the domain, when this problem does not exist. In the latter case, however, I would expect the question poser to have phrased it something like $(x,y)\to (1,1)$ in the domain $A=\Bbb{R}^2\setminus\{(x,y)\mid yx=\pm1\}$. $\endgroup$ – Jyrki Lahtonen Sep 5 '18 at 18:19
  • $\begingroup$ @JyrkiLahtonen Indeed I'm referring to the second definition that is the one I always refer to. The topic was recently discussed also here math.stackexchange.com/q/2889055/505767 and here math.stackexchange.com/questions/2902516/…. From that discussion I was inclined to consider your first definition as a simplified version used in hight school context but of course if you refer to that often there must be some reason for which in that context that definition is preferable. I will mention in future that point. $\endgroup$ – gimusi Sep 5 '18 at 19:01
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    $\begingroup$ Thanks for the links. Wikipedia agrees with the source I checked, but I am also inclined to put Rudin above WP any day of the week :-) $\endgroup$ – Jyrki Lahtonen Sep 5 '18 at 19:36
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This is right. Indeed the limit exists and it is 1/2.

Your line of reasoning is about right. A critical observation needed (which I think you used implicitly) is that for any $\epsilon$, there is a $\delta$ s.t. long as $||(x,y)-(1,1)|| \le \delta$, then $|1-\frac{1}{xy+1}| \le \epsilon$.

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Consider $X=x-1$ and $Y=y-1$, then we have $$ \begin{array}{c} \lim_{(x,y)\to (1,1)}\frac{xy-1}{x^2 y^2-1}= \lim_{(X,Y)\to (0,0)}\frac{XY-Y-X}{ \left( XY-Y-X+2 \right) \left( XY-Y-X \right)} =\lim_{(X,Y)\to (0,0)}\frac{1}{ \left( XY-Y-X+2 \right)}=\frac{1}{2} \end{array} $$

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    $\begingroup$ Why use this complicated way if we can cancel a factor also in the original expression? $\endgroup$ – gimusi Sep 5 '18 at 19:03
  • $\begingroup$ @gimusi It's a classical method , to solve these kind of limits when $(x,y)\not \to (0,0)$. For instance, consider $\lim \limits_{(x,y) \to (1,2)}\frac{xy-2x-y+2}{x^2 + y^2 - 2x -4y + 5}$. Now by selecting $x=X+1$ and $y=Y+2$ we get $\lim \limits_{(X,Y) \to (0,0)}\frac{X\,Y}{X^2+Y^2}$ which can be solved by Polar method. $\endgroup$ – user0410 Sep 5 '18 at 20:35
  • $\begingroup$ ok in some case it can be useful to reduce to limit at $(0,0)$ or $\infty$, it is a good trick to follow usually but in thatI can't see any simplification to use it. $\endgroup$ – gimusi Sep 5 '18 at 20:39
  • $\begingroup$ @gimusi After years of mathematical teaching of Calculus, I believed that the mentioned method is one of the best ways to solve these problems. In fact, the most powerful method for solving these kind of limits, is based on the the Polar system and for polar system we need to have $(x,y)\to (0,0)$. Therefore, I personally prefer that a student learn this method with a variety of examples, rather than just understanding a particular method. $\endgroup$ – user0410 Sep 5 '18 at 20:52
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    $\begingroup$ I also use polar coordinates often with multivariable limit and I agree with you that it is a foundamental method for that. Anyway really I think that for this particular case it is not a good application and you maybe should clarify that your aim is solely to show an application not effective here but in other cases. Of corse that only my point of view and it can be different from your one. Regards $\endgroup$ – gimusi Sep 5 '18 at 21:00

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