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In the first year calculus course at my university, we do not introduce the $\varepsilon$-$\delta$ definition of a limit. When considering the limit of a function of two variables, we resort to paths. That is, for the real-valued function $f$, defined in a deleted open neighbourhood of ${\bf a} \in \mathbb{R}^2$, we write $\displaystyle\lim_{{\bf x} \to {\bf a}}f({\bf x}) = L$ if the limit along every path to ${\bf a}$ is $L$.

Though we don't go into this detail, what is meant by the limit along a path is to pick a parameterisation $\gamma : (r, s) \to \mathbb{R}^2$ with $\displaystyle\lim_{t \to s^-}\gamma(t) = {\bf a}$, and then check that $\displaystyle\lim_{t \to s^-}f(\gamma(t)) = L$. Of course, without the $\varepsilon$-$\delta$ definition, this one-dimensional limit can only be deduced from limit laws and given fundamental limits.

Without loss of generality, suppose ${\bf a = 0}$. When asked to determine whether or not the limit $\displaystyle\lim_{{\bf x} \to {\bf 0}}f({\bf x})$ exists, there are only a couple of things the students can do:

  • employ continuity if possible (limit laws may be needed),
  • use polar coordinates together with the squeeze theorem,
  • find a path along which the limit does not exist, and
  • find two paths with have different limits.

The first two options can be used to show the limit exists, while the last two options can be used to show the limit does not exist. An efficient way to test limits along different paths is to try a whole family of paths simulateously, i.e. we could consider the family of quadratic paths given by $\gamma(t) = (t, kt^2)$ where $k \in \mathbb{R}$. If the limit $\displaystyle\lim_{t \to 0-}f(\gamma(t))$ exists and is independent of $k$, then along all of these paths, the limit is the same. This does not prove that $\displaystyle\lim_{{\bf x} \to {\bf 0}}f({\bf x})$ exists as there are still plenty of paths that have yet to be considered.


Questions:

Is there a reference which takes this approach to limits in higher dimensions (i.e. greater than one) in a rigorous way without using the $\varepsilon$-$\delta$ definition?

I wouldn't be surprised if the answer is no. However, I am somewhat more interested in the following question:

How pathological can two-dimensional limits be?

That is, along how many paths can the function have the same limit whilst still failing to have the overall limit existing? Is it possible to have all but one? I imagine there would be some sort of density argument that would prevent this.

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    $\begingroup$ I think using paths to define limits in higher dimensional spaces is sick. In this way you have to test a gogol of parametrized paths converging to ${\bf a}$ in order to prove a single instance of $\lim_{{\bf x}\to{\bf a}}f({\bf x})=L$. $\endgroup$ – Christian Blatter Aug 5 '13 at 8:43
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    $\begingroup$ Michael, you can, for example, given any $K>0$, define a function that has limit $0$ along all paths $y=cx^k$ for $0<k\le K$ but nevertheless has no limit at $0$. Here's an exercise for you: Prove that if $\lim_{t\to 0} f(g(t))=\ell$ for all continuous paths $g$ with $g(0)=0$, then $\lim_{x\to 0} f(x)=\ell$. $\endgroup$ – Ted Shifrin Aug 5 '13 at 9:46
  • $\begingroup$ I agree with Christian, and even more if that is the first approach you people do towards limits...! I think I good deal if mathematical intuition is needed to get into several variables calculus, and you begin with limits of two variable functions? It doesn't look good to me... $\endgroup$ – DonAntonio Aug 5 '13 at 10:32
  • $\begingroup$ Just to be clear, they don't get told anything about parameterisations, I was just explaining what was meant. The two-dimensional limits section is just tacked on at the end of the course. They do it properly in the multivariable calculus course. Furthermore, I don't get to decide what is taught; if I did, I would get rid of the multidimensional limits altogether. $\endgroup$ – Michael Albanese Aug 5 '13 at 12:30
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An observation to address your second question: If there exists two paths that approach $0$ which give different values for $f$, then there exist infinitely many paths approaching $0$ for which $f$ has no limit.

Suppose $\gamma_1(t)$ and $\gamma_2(t)$ give $f$ different limits as $t$ approaches $0$ from the left. Then for any real $c\ne0$, consider the path

$$\gamma_c(t)=\sin\left(\frac{c}{t}\right)\gamma_1(t)+\cos\left(\frac{c}{t}\right)\gamma_2(t)$$

To see that $\lim_{t\to 0^-}f(\gamma_c(t))$ does not exist, consider the subsequence $t_n=-\frac{2c}{\pi n}$.

However, if we only consider paths whose limits exist then we can construct a function in which there is only one deviant path. Let

$$f(x,y)=\begin{cases}1&y\ne 0\text{ or }x\ge0\\0&y=0\text{ and }x<0\end{cases}$$

Notice that any path whose limit exists will give a limit of $1$, except if there is a neighborhood around the origin in which the path approaches along the $x$-axis from the left. So, ignoring reparametrizations, and considering two paths equal if their images agree in some neighborhood of the origin, we have defined a function with a unique deviant path.

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