10
$\begingroup$

Given the function $f(x,y) = \frac{xy}{x+y}$, after my analysis I concluded that the limit at $(0,0)$ does not exists.

In short, if we approach to $(0,0)$ through the parabola $y = -x^2 -x$ and $y = x^2 - x$ we find that $f(x,y)$ approaches to $1$ and $-1$ respectively. Therefore the limit does not exists.

I think my rationale is right. What do you think?

Alternatively, is there another approach for this problem?

  • 4
    Your approach is correct. The limit must be independent of the path to the origin.2012-03-14
  • 0
    $f(x,y)$ can not be determine at $(0,0)$2012-03-14
  • 0
    Note that your paths are tangent at the origin to the line $y=-x$, which is not in the function's domain. When looking for paths for limit calculations like this, finding such tangent paths is one of the strategies. (Another basic strategy, which is not helpful in this problem, is to try $y=kx$.)2012-03-18
  • 0
    @alex.jordan A selection of a few examples where tangent paths are the way to go would make for a nice post on the site.2018-03-19

5 Answers 5

6

I'll explain here how to approach limits of functions in two variables, with the example the OP proposed in mind. If the limit $$\lim_{(x,y)\to (0,0)} \frac{xy}{x+y}$$ exists and equals $L$, then it also follows that if $\{(x_n,y_n)\}$ is a sequence of points with limit $(0,0)$, then $$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=L.$$ Now we can choose a number of easy sequences $\{(x_n,y_n)\}$ with limit $(0,0)$, and calculate the limit. For instance, we can pick points in a line $y=\lambda x$, with slope $\lambda$, i.e., $(x_n,y_n) = (\frac{1}{n}, \frac{\lambda}{n})$. In this case: $$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{\lambda}{n^2}}{\frac{1}{n}+\frac{\lambda}{n}}=\lim_{n\to\infty} \frac{\lambda}{(1+\lambda)n}$$ and the limit is $0$ as long as $\lambda\neq -1$. Hence, if the limit exists, it must be $0$. But the problem with $\lambda=-1$ tells us that there may be a problem if we approach $(0,0)$ with a path that ends tangent to $y=-x$ (notice that the function is not defined at points with $y=-x$).

Thus, next we look at a sequence following a path on a curve with tangent line $y=-x$ at $(0,0)$. Examples of such curves include $y=x^2-x$, $y=-x^2-x$ or $y=e^{-x}-1$. Thus, we may consider sequences $(x_n,y_n)$ given by: $$\left(\frac{1}{n},\frac{1}{n^2}-\frac{1}{n}\right),\quad \text{or} \quad \left(\frac{1}{n},-\frac{1}{n^2}-\frac{1}{n}\right), \quad \text{or} \quad \left(\frac{1}{n},e^{-1/n}-1\right).$$ For the first sequence we obtain: $$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{1}{n^3}-\frac{1}{n^2}}{\frac{1}{n}+\frac{1}{n^2}-\frac{1}{n}}=\lim_{n\to\infty} \frac{\frac{1}{n^3}-\frac{1}{n^2}}{\frac{1}{n^2}}=\lim_{n\to\infty} \frac{1-n}{n}= -1.$$ But the limit was supposed to be $L=0$. Hence the limit cannot exist. Similarly, if we try the other two sequences listed above: $$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{-\frac{1}{n^3}-\frac{1}{n^2}}{\frac{1}{n}-\frac{1}{n^2}-\frac{1}{n}}=\lim_{n\to\infty} \frac{-\frac{1}{n^3}-\frac{1}{n^2}}{-\frac{1}{n^2}}=\lim_{n\to\infty} \frac{1+n}{n}= 1,$$ and $$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{n(e^{-1/n}-1)}{n+e^{-1/n}-1}=\lim_{n\to\infty} \frac{e^{-1/n}-1}{1+\frac{e^{-1/n}}{n}-\frac{1}{n}}=-1.$$ These results are inconsistent, and therefore the limit cannot exist. Even more dramatic: let $\{x_n,y_n\}$ be a sequence following the curve $y=x^3-x$ towards the origin, for instance put $(x_n,y_n)=(\frac{1}{n},\frac{1}{n^3}-\frac{1}{n})$. Then: $$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{1}{n^4}-\frac{1}{n^2}}{\frac{1}{n}+\frac{1}{n^3}-\frac{1}{n}}=\lim_{n\to\infty} \frac{\frac{1}{n^4}-\frac{1}{n^2}}{\frac{1}{n^3}}=\lim_{n\to\infty} \frac{1-n^2}{n}= -\infty.$$

1

The key point is to consider approaching the origin near the line $y = -x$. No matter how small a neighborhood of the origin you consider, in that neighborhood $xy/(x+y)$ takes on every value. See the plot of $xy/(x+y)$.

You could also assume the limit exists and, using the definition of the limit of a multivariable function (with epsilons and deltas), arrive at a contradiction.

  • 0
    The line $y=-x$ is not in the domain of the function.2012-03-14
  • 0
    On the other hand, the fact that $f(x,y)$ blows up near the line $y=-x$, no matter how close $(x,y)$ is to $(0,0)$, is another way to see that the limit cannot exist.2012-03-14
  • 0
    @Jonas: yes, $y = -x$ is not in the domain, but isn't this an (admittedly somewhat legalistic) way to show that the limit doesn't exist? Usually at the calculus level we require the function to be defined in some deleted neighborhood of a point in order to contemplate the limit at that point. (But your observation that the singularities are not removable is a better way to go...)2012-03-14
  • 0
    @Pete: Probably. At least in this case there is no limit regardless, and I agree that it's best that both legalistic and other reasons are considered to see what's going on. A strictly legalistic example like $f(x,y)=\frac{xy}{x}$ would be less interesting.2012-03-14
  • 0
    @Pete: I apologize for my ignorance, what do you mean by "legalistic"? In my mind the simplest proof is the best. Is there a simpler proof?2012-03-14
  • 0
    @oenamen: consider the example $\frac{xy}{x}$. This function is essentially just $y$, except for the fact that it's written in such a way that it is undefined when $x=0$. Since it is undefined at points arbitrarily close to $(0,0)$ (along the line $x=0$), one might say that its limit doesn't exist at $(0,0)$. However, this is not a property of the function itself being badly behaved near $(0,0)$ in terms of its values, only a result of the domain having been restricted. In some sense the singularities are removable: the function can be continuously extended to...2012-03-14
  • 0
    ...the entire plane by the formula $y$. If $G$ is a region in the plane and $p$ is a limit point of $G$, then even if $G$ does not contain a deleted neighborhood of $p$, one may consider things like $\lim\limits_{(x,y)\to p, (x,y)\in G}f(x,y)$, limits of functions defined on $G$, only considering values at points in $G$. For $G=\{(x,y):x\neq 0\}$, we have $\lim\limits_{(x,y)\to(0,0),(x,y)\in G}\frac{xy}{x}=0$. But if we set up the "law" that limits are only defined when the domain contains the entire deleted neighborhood, then the limit doesn't exist.2012-03-14
  • 0
    In the question here there are other reasons the limit doesn't exist. Since in some contexts this domain restriction isn't essential, it could in some contexts be seen as a "legalistic" reason for the limit not to exist. That doesn't take away its validity here. (Sorry for being so long winded.)2012-03-14
  • 0
    @JonasMeyer: Thanks for the clarification. Perhaps I should write "It is simpler to consider approaching the origin near the line $y=-x$."2012-03-14
  • 1
    @oenamen: I agree. Actually, that is basically what Rolando did, but making more explicit that that is the key point is helpful.2012-03-14
1

That is a good approach.

Generalizing slightly, if $g(x)$ is a function defined in a punctured neighborhood of $0$ such that $\lim\limits_{x\to 0} g(x) = 0$ and $g(x)\neq 0$ for all $x$, then if the limit in question exists, you should have $\lim\limits_{(x,y)\to (0,0)}f(x,y)=\lim\limits_{x\to 0}f(x,g(x)-x)=\lim\limits_{x\to 0}x -\frac{x^2}{g(x)}$. Such $g$ can be chosen to make this limit be any real number, $\infty$ or $-\infty$, or not exist in any sense. Another particularly easy special path is to consider $\lim\limits_{x\to 0}f(x,0)=0$.

0

Consider the path $y=1-e^x$. Then $$...=\lim_{x\to0}\frac{x(1-e^x)}{x+(1-e^x)}=-2$$

0

maybe I miss something but isn't it:

for $x=x,y=\dfrac{-x}{x+1}$

$\dfrac{xy}{x+y}=\dfrac{\dfrac{-x^2}{1+x}}{x-\dfrac{x}{1+x}}=\dfrac{\dfrac{-x^2}{x+1}}{\dfrac{x^2}{x+1}}=-1\to0$ when $(x,y)\to(0,0)$

and for $x\not=0,y=0$

$\dfrac{xy}{x+y}=0\to0$