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For example,

$\displaystyle \lim_{(x,y,z)\rightarrow(0,0,0)} \frac{x^2y^2z^2}{x^2+y^2+z^2}$

This question is from 8th edition of Stewart Calculus textbook. My fellow graduate student TAs and myself recently got many students ask this question in office hour/by email.

Because this is a freshman level calculus class, most instructors choose to only briefly explain this topic and (probably) do not expect students to write a full proof of such a problem on the exams. In our single variable calculus course syllabus the $\epsilon-\delta$ definition of limit is also omitted.

My question is, for a few people who do want to know how to solve this problem, what's the best way to explain?

Some thought

  1. Convince them they must know the rigorous limit definition for such a problem. (Any better approach?)
  2. Bound the function by 0 and $(x^2+y^2+z^2)^2$ and apply squeeze theorem. (Almost as hard as 1.?)
  3. Change to spherical coordinate. (They won't learn this until they get to the multivariable integration chapter.)
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    $\begingroup$ For a student who hasn't had the rigorous $\epsilon - \delta$ definition (or who has merely been exposed to it, but not enough to really work with it), option 2 is probably the best (and is what Stewart does, if I recall correctly). Also, even though it isn't a proof, you can show that on all lines through the origin the corresponding 1-dimensional limit is zero. This could perhaps build intuition, at the possible risk of students believing that the argument is in and of itself an adequate proof. $\endgroup$ – John Coleman May 22 '17 at 10:58
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    $\begingroup$ They might be familiar/comfortable with polar coordinates, so conceptually spherical coordinates might not be too much of a leap. $\endgroup$ – john w. May 23 '17 at 14:40
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Showing why there is a need for good definitions here is probably more important than showing students why a difficult definition "works" via plug-in hocus-pocus. (Though if you have the time to do the proper definition properly that is even better; I certainly don't in multivariate calc. Squeeze theorem, as commenter John Coleman points out, is a good compromise there, as long as you make it clear you know it's a deus ex machina for them at this point.)

With modern software this becomes very easy. You can do it with just about anything; I use SageMath, and also find CalcPlot3D a good (free) option.

Here is what I use in this situation (modifying your example slightly in the first instance since visualizing yours would require students "seeing" how contour surfaces work). You can surely use whatever single graphic is in the text and modify it in your language of choice.

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You can make this $\epsilon/\delta$ proof easy enough that an interested student should find the argument believable without experience with $\epsilon/\delta$ proofs. Divide both the numerator and denominator by $x^2 y^2 z^2$ to get $$\frac{1}{ \frac{1}{y^2 z^2} + \frac{1}{x^2 z^2} + \frac{1}{y^2 z^2}}$$ Then if $|x|,|y|,|z| < \epsilon$ then we have $$ \frac{1}{y^2 z^2} + \frac{1}{x^2 z^2} + \frac{1}{y^2 z^2} > \frac{3}{\epsilon^4}.$$ Thus $$\frac{1}{ \frac{1}{y^2 z^2} + \frac{1}{x^2 z^2} + \frac{1}{y^2 z^2}} < \frac{\epsilon^4}{3}$$

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$\displaystyle \lim_{(x,y,z)\rightarrow(0,0,0)} \frac{x^2y^2z^2}{x^2+y^2+z^2}$

Even if this were a course where students were required to write epsilon-delta proofs, I would still imagine that anyone competent would approach this problem by first thinking informally about why this limit is expected to exist and be zero. I would split this into three levels of rigor:

(1) As you approach the origin, each of the three coordinates gets very small. The numerator is sixth order and the denominator second order, so we expect the numerator to get small much faster than the denominator. The first goal should be to get students to the level where this mode of reasoning seems natural to them.

(2) Next, I would worry about whether this argument fails for some special path leading to the origin. The only obvious possibilities that might have such "special" behavior are paths lying along the axes -- these are the only paths that seem obviously to have been singled out by the structure of the expression. We check that the argument in #1 still holds for these paths.

(3) Do something more rigorous, such as the possibilities you suggested in the question.

Given the level at which your course is being taught, #3 doesn't seem like a realistic expectation on a homework problem.

I think it is more important for students to be competent enough to do #1 than for them to be able to cook up some tricky approach that applies specifically to this problem.

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Let's consider a path $$\left( x(t), y(t), z(t) \right)$$ with $\left( x(0), y(0), z(0) \right) = (0,0,0)$. For small $t$, we can approximate this path with a straight line, in which case $$\left( x(t), y(t), z(t) \right) = (at, bt, ct)$$ for some constants $a,b,c$. This is admittedly non-rigorous, but it should seem intuitively plausible that a smooth curved path looks straight if you zoom in close enough -- that, after all, is a fundamental idea in Calculus 1.

So now substitute that into the formula: We need to compute $$\lim_{t \to 0} \frac{(at)^2(bt)^2(ct)^2}{(at)^2+(bt)^2+(ct)^2}$$ which is a limit they can calculate by single-variable methods.

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    $\begingroup$ While I think it is good to discuss this idea, one should also point out that it doesn't always work: there are functions whose limit exists along every line passing through a point, but where the limit at that point does not exist. $\endgroup$ – Steven Gubkin May 23 '17 at 3:07
  • $\begingroup$ @StevenGubkin Yes, that's certainly true. The condition you really need is not "along every line", but "along every path". Unfortunately the latter does not lend itself well to computation. $\endgroup$ – mweiss May 23 '17 at 3:23
  • $\begingroup$ @StevenGubkin However, if one is trying to do this without the epsilon-delta definition, one's hands are somewhat tied. $\endgroup$ – mweiss May 23 '17 at 3:23
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    $\begingroup$ Sure. But there are some simple examples which can convey the problem. For instance, $f(x,y)=0$ if $y=x^2$ with $x \neq 0$, and $f(x,y) = 1$ otherwise. This "clearly" has one limit along each line, and a different limit along the parabola. $\endgroup$ – Steven Gubkin May 23 '17 at 3:27

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