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A couple months ago I had a math test which I couldn't do this two-part exercise,

Given $f(x,y)=\frac{(x-1)^2(y-1)}{(x-1)^4+(y-1)^2}$ and $g(x,y)=\frac{(x-1)^2(y-1)^2}{(x-1)^4+(y-1)^2}$

So the question for both parts was find, if it exists, the limit as $(x,y)\rightarrow(1,1)$

Today after reading a little about methods for calculating limits I thought of the following solution:

1) $0\le\lvert f(x,y)\rvert = \frac{(x-1)^2\lvert(y-1)\rvert}{(x-1)^4+(y-1)^2}$

2) $(x-1^2)\le (x-1)^4 + (y-1)^2 \rightarrow \frac{(x-1)^2}{(x-1)^4+(y-1)^2} \le 1 \rightarrow \frac{(x-1)^2}{(x-1)^4+(y-1)^2}*\lvert(y-1)\rvert \le \lvert (y-1)\lvert$

Therefor $\lim \limits_{x,y \to 1,1}0\le \lim \limits_{x,y \to 1,1}\frac{(x-1)^2\lvert(y-1)\rvert}{(x-1)^4+(y-1)^2} \le \lim \limits_{x,y \to 1,1}\lvert (y-1)\lvert$ $\rightarrow 0 \le \lim \limits_{x,y \to 1,1}\frac{(x-1)^2\lvert(y-1)\rvert}{(x-1)^4+(y-1)^2} \le 0$

So $\lim \limits_{x,y \to 1,1}|f(x,y)| = 0 \rightarrow \lim \limits_{x,y \to 1,1}f(x,y) = 0$

And then again the exact same proof for $g(x,y)$ since the only difference is on $(y-1)$ being squared on the second function. So my question is, is this correct? I mean, it seems weird to me to have two exercises in a single exam with exact same solution

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Neither method presented here is actually correct. The limit of the first function does not exist. I will also shift my variables and look at $$ h(x,y) = \frac{x^2y}{x^4 + y^2}. $$ First, let's look at $y = x$, we have $$ \lim_{x \to 0} h(x,x) = \lim_{x \to 0} \frac{x^3}{x^4 + x^2} = \lim_{x \to 0} \frac{x}{1+ x^2} = 0 .$$ On the other hand, look at $x^2 = y$. Then, $$ \lim_{y \to 0} h(x, x^2) = \lim_{x \to 0} \frac{x^4}{2x^4} = \frac{1}{2}. $$ Since these do not agree, the limit does not exist. It might be illuminating for you to plot the graph of this function.

For the second one, you can show that the limit is zero from the squeeze theorem.

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  • $\begingroup$ Wow that is actually pretty interesting and it seems completely logical. So both methods only show the limit if it exists? As in if the limit doesn't exist like this case, the methods presented will provide a wrong answer? $\endgroup$ – Chapi Feb 19 '16 at 0:56
  • $\begingroup$ Well, both methods should work if you are careful. When using polar coordinates, you can see from this example that it is not enough that the limit is zero for every $\theta$. You really need to finish the argument still with a squeeze theorem, but polar coordinates can help to set this up. For your method, it should work, but the inequality $(x-1)^2 \leq (x-1)^4 + (y-1)^2$ is not true. (take $x = 3/2, y = 0$) $\endgroup$ – Andrew Hanlon Feb 19 '16 at 5:40
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    $\begingroup$ I meant to write $y=1$ $\endgroup$ – Andrew Hanlon Feb 19 '16 at 5:48
  • $\begingroup$ I think I understand now, and yes I noticed that the limit isn't zero for every angle $θ$ when using polar coordinates. So another thing that I'm wondering now, if you use polar coordinates over a function to calculate the limit and you find out that the limit does not exist because it depends on $θ$, is that enough proof to show that the limit does not exist? And in case it shows that the limit does not depend on $θ$ is that enough to say it does exist, or do I still need to further argument using the squeeze theorem? Thanks for the very helpful answers by the way. $\endgroup$ – Chapi Feb 19 '16 at 15:11
  • $\begingroup$ If you see that the limit depends on $\theta$, then it is enough to show that the limit does not exist (if it did, no matter what angle you approach the origin from, you will get the same answer). If the limit does not depend on $\theta$, it is not enough. As you can see from this problem, it is possible for the limit to approach zero along every line to the origin, but still not exist. $\endgroup$ – Andrew Hanlon Feb 20 '16 at 1:02
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It is easier to change the function to $h(x,y) = f(x+1, y+1)$ and limit to $(x,y) \to (0,0)$ and then use polar coordinates:

$$\lim_{(x,y)\to (0,0)} {x^2y\over x^4+y^2} = \lim_{r \to 0} {r^3 \cos^2(\theta)\sin(\theta)\over r^2(r^2 \cos^4(\theta) + \sin^2(\theta))} = \lim_{r \to 0} {r \cos^2(\theta)\sin(\theta)\over r^2 \cos^4(\theta) + \sin^2(\theta)} = 0$$

Since $\lim_{(x,y)\to (0,0)}h(x,y) = 0$, also $\lim_{(x,y)\to (1,1)}f(x,y) = 0$

Analogously proceed for $g(x,y)$.

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  • $\begingroup$ Well that's a great method I did not know about and yeah it seems much simpler, although it still doesn't answer my question which is if the method I posted is correct or not, thanks anyways! $\endgroup$ – Chapi Feb 13 '16 at 20:37

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