# Multivariable limits

Multivariable limits are harder than their one-variable counterparts, and textbooks examples usually focus on limits that don't exist when approaching from different straight lines. This gives the false impression that testing straight lines is "the method" to use when trying to discover if a given limit exists. Counterexamples as $$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^4+y^2}$$ don't convince students. The question is: what are effective strategies to convince them that there is no single way to attack these problems?

• Ask them to produce their own counterexamples? Also, I find piecewise defined functions much easier to understand why this behavior occurs, like $f(x,y) = \begin{cases}0 \text{ if } y = x^2 \\ 1 \text{ else}\end{cases}$ – Steven Gubkin Apr 22 '14 at 13:27
• If you look about half way down supermath.info/ZooOfMathematicalCreatures.html you'll see where I animate a few of the standard examples where various paths reveal different behavior. – James S. Cook Apr 23 '14 at 2:01

A problem I used to assign on take-home tests back when I was teaching gifted high school students (late 1990s) might be of interest.

Show that $f(x,y) =y^{x}$ has the same limit along any 1st quadrant polynomial approach towards $(0,0)$ (i.e. along any path having the form $y=ax^{n},$ where $x>0$ and $n$ is a positive integer), but the limit varies if approach paths of the form $y=\exp \left(-\frac{a}{x}\right)$ are used, where $a>0.$ (Why do I need $a>0?)$ As part of your solution include carefully drawn graphs, for $0<x<0.5,$ of $y=x^{2}$, $y=x^{3},$ $y=x^{4},$ $y=\exp \left(-\frac{0.5}{x}\right),$ $y=\exp \left(-\frac{1}{x}\right),$ and $y=\exp \left(-\frac{2}{x}\right)$ together on the same coordinate axis. You may sacrifice some accuracy in order to give a more informative and schematic presentation of the behavior of these graphs as $x \rightarrow 0^{+}.$

(Extra Credit) Verify analytically that if $a>0$ and $n$ is any positive integer, then $\lim\limits_{x\rightarrow 0^{+}}\left[ \frac{\exp \left(-ax^{-1}\right)}{x^{n}}\right] =0.$ You will find that a direct application of L'Hopital's rule doesn't help. However, this limit can be evaluated by using the changing variables method given (somewhere) in my handwritten handout titled "Some Techniques for Evaluating Limits". But feel free to use another (mathematically justified) method if you find one.

Note: The problem itself can be solved easily by noting the following.

$$y^x \;\; = \;\; \left(ax^{n}\right)^x \;\; = \;\; a^{x} \cdot x^{nx} \;\; = \;\; a^{x} \cdot \left(x^x\right)^n \;\; \rightarrow \;\; a^0 \cdot 1^n \;\; = \;\; 1$$

$$y^x \;\; = \;\; \left(e^{-\frac{a}{x}}\right)^x \;\; = \;\; e^{-a} \;\; \rightarrow \;\; e^{-a}$$

The point of the extra credit is to show analytically what the graphs should show, namely that the exponential approaches slip underneath every one of the graphs $y = ax^n$ as $x \rightarrow 0^{+}.$

I think the whole problem is connected to the problem of quantifiers, like "for all" and "there is". See also this question which I feel is relevant here.

The important point is to stress that the definition of the limit ($\varepsilon$-$\delta$ or sequences, does not really matter) means that you have to get the same limit when approaching the point independently of the way how you approach it.

To prove that a limit does not exist, it is sufficient to show that there are two possible ways to get close and get different limits, and the easiest possible way is to do this is by checking straight lines.

But to show existence, you have to prove it for every possible approach. This makes these exercises far more difficult if the limit exists, because to prove the converse it is sufficient to see one example, but to have the limit, you have to have it all.

Examples like the one you mention and also in other answers are great to stress that if you do not find a counterexample it does not mean there is no.

• I think this is extremely insightful, in that it recognizes the asymmetric nature of the two different tasks ("Prove the limit exists" & "Prove the limit does not exist"). – mweiss Apr 25 '14 at 1:21

I'd go by the definition (an $\delta$ neighborhood of $(0, 0)$ in your case) a few times. Perhaps cook up some limit that is easy with straight lines in a different coordinate system (but don't tell them the weird curves you choose are straight there until the end!).

It is always useful to give a homework comparing several approaches, e.g. prove the limit using the definition, use straight lines in all directions (essentially polar coordinates), go by a spiral, ... If for each way you can find problems that are easy to solve and hard in the others, you'll probably get what you want. Do mention them here ;-)

This idea is not (yet) supported by experience, but one thought that occurs to me is to emphasize the "infinitesimal" definition of limit: $\lim_{(x,y)\to(0,0)} f(x,y) = L$ if $f(x,y)$ is infinitely close to $L$ whenever $x$ and $y$ are both infinitely close to $0$ (but not both equal to it), i.e. whenever $(x,y)$ is in a punctured two-dimensional "infinitesimal region" around $(0,0)$. The point is then that the ratio of two infinitesimals can itself be infinitesimal or infinite. E.g. for $f(x,y) = \frac{x^2y}{x^4+y^2}$, if $\epsilon$ is infinitesimal, then so is $\epsilon^2$, but we have $f(\epsilon,\epsilon^2) = \frac{1}{2}$ which is not infinitesimal.

One could avoid mentioning infinitesimals by instead talking about "small" or "close" numbers.