Process of finding limits for multivariable functions

Question

I was tutoring a student today and they asked a question which made me curious.

We were working on the following question together.

After explaining that we must look at the limit along the x axis, I mentioned that we then need need to look at the limit along the path y=mx.

We got to the step where we would find f(x,xm) but when I looked at his paper he had f(y/m,xm). Now, he quickly established that he ended up going in circles however, when he asked if there was a mathematical reason why this did not work, I couldn’t tell him.

So, is there, if any, a mathematical reason why we only change the y value and keep the x value as is?

Answer 1

There's nothing wrong with "changing" or substituting expressions for both $x$ and $y$. However, we can only use one variable across both expressions.

What does looking at a function along different paths mean? For our purposes, a path is just a curve, which we often associate with a parametrization $\vec{r}(t)$. So if we have a two-variable function $f(x,y)$, when we restrict our attention to a specific path, what we're really doing is considering $f\bigl(\vec{r}(t)\bigr)$, which is a single-variable function. This allows us to use all of our tools from single-variable calculus!

I think many textbooks obscure this idea in their treatments of multivariable limits by presenting evaluating a function along a path as performing a substitution. For example, following a certain popular commercial text, the problem you gave would be done like this.

Along the $x$-axis, every point is of the form $(x,0)$, and the limit along this approach is $$\lim_{(x,0)\to(0,0)}\frac{x \cdot 0}{x^2+0^2}=0.$$ However, when $(x,y)$ approaches $(0,0)$ along the line $y=x$, we obtain $$\lim_{(x,x)\to(0,0)}\frac{x \cdot x}{x^2+x^2}=\frac{1}{2}.$$ The limit does not exist because two approaches produce different limits.

Another popular text is similar, perhaps marginally better.

Let's approach $(0,0)$ along the $x$-axis. On this path $y=0$ for every point $(x,y)$, so for all $x\neq0$, the function becomes $$f(x,0)=0/x^2=0,$$ and thus $$f(x,y)\to0 \quad \text{as} \quad (x,y)\to(0,0)\,\text{ along the $x$-axis.}$$ Let's now approach $(0,0)$ along another line, say $y=x$. For all $x\neq0$, $$f(x,x)=\frac{x^2}{x^2+x^2}=\frac{1}{2},$$ therefore $$f(x,y)\to\tfrac{1}{2} \quad \text{as} \quad (x,y)\to(0,0)\,\text{ along $y=x$.}$$ Since we have obtained different limits along different paths, the given limit does not exist.

In both of these approaches, it appears as though we're simply performing the substitutions $(x,y)=(x,0)$ and $(x,y)=(x,x)$. So why doesn't $(x,y)=(y,x)$ work? Or if we want to approach $(0,0)$ along $y=2x$, why can't we do $(x,y)=(y/2,2x)$? It's because these aren't parametrizations of curves—they aren't paths.

Vector-valued functions and parametrizations of curves are presented before multivariable limits in both of the texts I referenced. So I think a better approach would be something like this.

We can approach $(0,0)$ along the $x$-axis by following the path $$\vec{r}(t)=(t,0)\,\text{ as }\, t\to0.$$ This gives us $$\lim_{t\to0} f\bigl(\vec{r}(t)\bigr)=\lim_{t\to0}\frac{t \cdot 0}{t^2+0^2}=0.$$ We can also approach $(0,0)$ along the line $y=x$ by following the path $$\vec{s}(t)=(t,t)\,\text{ as }\, t\to0.$$ This gives us $$\lim_{t\to0} f\bigl(\vec{s}(t)\bigr)=\lim_{t\to0}\frac{t \cdot t}{t^2+t^2}= \frac{1}{2}.$$ The limits along different paths do not agree, therefore the limit does not exist.

Answer 2

Presumably your textbook has some theorem along the lines of the following, which sets up the goal of trying different paths:

Theorem. If $f$ is a real-valued function such that $$\lim_{(x,y)\rightarrow(a,b)} f(x,y) = L \,,$$ then for all real numbers $p$, $q$ not both zero, $$\lim_{t\rightarrow0} f(a+pt,b+qt) = L \,.$$

Instead of the parametrized lines $(a+pt,b+qt)$, the theorem might be about more general paths ${\bf r}(t)$ that approach $(a,b)$ as $t$ approaches $c$ but do not pass through $(a,b)$ for $t\ne c$. (Or perhaps your textbook relies only on the uniqueness of the limit, which would complicate the contrapositive argument below.)

One then applies the contrapositive of the theorem (and maybe this is the relevant theorem in your textbook): If you get different one-variable limits along different paths through $(a,b)$, then the two-variable limit does not exist.

Whatever the statement of the theorem, the goal is to find one-variable limits that disagree; then you win. If the limits agree, then you lose: find another path or reconsider whether the two-variable limit exists.

With that goal in mind, $f(y/m, mx)$ is a bad problem-solving move, since there are still two variables. The student missed the goal of getting down to one variable. Either $f(y/m, y)$ or $f(x, mx)$ is a good move, however. Or you can change both $x$ and $y$ in terms of the one variable $t$, which is the way I prefer to teach it. Even $x=0, y=t$ instead of just $x=0$. I'd accept either, but I always present a solution to students in parametric form.

P.S. In polar coordinates, the expression becomes $\cos\theta\sin\theta$, and the one-variable limits depends on the angle of approach. Just another way to view the limit.