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Background

In multivariable calculus/complex analysis, one often tries to show that a limit doesn't exist by finding two paths which yield different limits. The justification goes that if the limit exists, it must be the same along any path; hence if it is not, it doesn't exist.

On an intuitive level, I think I understand this concept. But I have never seen a rigorous justification for this method, nor what exactly constitutes a "path".

Attempt at a solution

Definition/Theorem

I'm going to stick to $\mathbb{R}^2$ here, since the arguments should generalize easily.

Let $f: U \subseteq \mathbb{R}^2 \to \mathbb{R}$ be a real-valued function defined on the domain $U$, an open proper subset of $\mathbb{R}^2$. Let us define a "path" in this context to be a continuous function $\gamma: \mathbb{R} \to \mathbb{R}^2$, parameterized by $t$.

Suppose that $\lim\limits_{\mathbf{x} \to \mathbf{x_0}} f(\mathbf{x_0})$ exists. Then, for any path $\gamma$ with range in $U$ and $\gamma(t_0) = \mathbf{x_0}$,

$$\lim\limits_{t \to t_0} f(\gamma(t_0)) = L. $$

In terms of $\delta$'s and $\epsilon$'s, this says that for any $\epsilon > 0$, there exists some $\delta > 0$ such that

$$\left|t - t_0 \right| < \delta \Rightarrow \left| f(\gamma(t)) - f(\gamma(t_0)) \right| < \epsilon. $$

Proof

I don't know how to prove this.

Example

To demonstrate, I will show that the limit in example (d) from Paul Dawkins' Calculus III notes doesn't exist. I'll follow his proof, exchanging his intuitive definition with the formal one above.

To show: $\lim\limits_{(x, y) \to (0, 0)} \frac{x^3y}{x^6 + x^2}$ doesn't exist.

Proof: Suppose that the limit exists and is equal to $L$.

First, take the continuous path $\gamma(t) = (t, t)$ (the path $y = x$). By our theorem, we conclude that $$\lim\limits_{t \to 0} \frac{t^3t}{t^6 + t^2} = \lim\limits_{t \to 0} \frac{1}{t^2 + 1} = 0 = L.$$

But instead take the path $\gamma(t) = (t, t^3)$ (the path $y = x^3$). By our theorem, we conclude that this is also equal to $L = 0$, but we have $$\lim\limits_{t \to 0} \frac{t^3(t^3)}{t^6 + (t^3)^2} = \lim\limits_{t \to 0} \frac{1}{2t^6} = \frac{1}{2},$$ which is a contradiction. Thus, the limit does not exist.

Summary

So the definition seems to work. My questions are (1) is the definition/theorem I've stated above in terms of continuous paths the most general one and (2) how can we prove it?

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Sticking to paths is a needless (but useful) particularization of the following.

Definition: Let $f$ and $U$ be as in the question and let $A\subseteq U$ with $A\neq \varnothing$ and $\mathbf{x_0}\in \overline {A\,}$. Define $\lim \limits_{\mathop{\mathbf{x}\to \mathbf{x_0}}_{\large{\mathbf{x}\in A}}}\left(f\mathbf({x})\right):=\lim \limits_{\mathbf{x}\to \mathbf{x_0}}\left(f_{|A}(\mathbf{x})\right)$. The LHS is called a sublimit of $f$ at $\mathbf{x_0}$.

Claim: If $\lim \limits_{\mathbf{x}\to \mathbf{x_0}}\left(f(\mathbf{x})\right)$ exists and equals $\mathbf {y_0}$ for some $\mathbf {y_0}\in \mathbb R$, then for all $A\subseteq U$, $\lim \limits_{\mathop{\mathbf{x}\to \mathbf{x_0}}_{\large{\mathbf{x}\in A}}}\left(f\mathbf({x})\right)$ exists and equals $\mathbf{y_0}$.

Proof: It's immediate. The same $\delta$ that works for $\lim \limits_{\mathbf{x}\to \mathbf{x_0}}\left(f(\mathbf{x})\right)$, trivially works for $\lim \limits_{\mathop{\mathbf{x}\to \mathbf{x_0}}_{\large{\mathbf{x}\in A}}}\left(f\mathbf({x})\right)$ because $A\subseteq U$. $\square$

For the particular example of paths, just let $A=\text{im}(\gamma)$.

What's useful it the contrapositive of the claim whichs claims that if there exist $A_1, A_2\subseteq U$ such that the sublimits differ (or do not exist), then the limit doesn't exist.

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  • $\begingroup$ Ah, so much simpler than I had imagined! I wonder why this is not mentioned more often in textbooks, since it is so simple. Thanks! $\endgroup$ – AmadeusDrZaius Jul 10 '14 at 16:42
  • $\begingroup$ @AmadeusDrZaius My experience with multivariable calculus literature is null (I just know my course notes). I find this very important to mention. $\endgroup$ – Git Gud Jul 10 '14 at 16:46
  • $\begingroup$ Theoretically this works for an arbitrary subdomain $A$, so I guess the issue one runs into is just that $\lim\limits_{\mathbf{x} \to \mathbf{x_0}} \left( f_{|A}(\mathbf{x})\right)$ may not exist depending on how $A$ is chosen? E.g., the limit exists for both $A = U \cap \mathbb{Q}$ (the rational points in $U$) and $A = \{(x, y) \in \mathbb{R} \mid y = x^3$ (a continuous path)? $\endgroup$ – AmadeusDrZaius Jul 10 '14 at 16:53
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    $\begingroup$ @AmadeusDrZaius The answer to the question is your comment is yes. But I wouldn't say it is 'the' issue. $\endgroup$ – Git Gud Jul 10 '14 at 17:12
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A more literal proof.

The missing ingredient is convergence in $\mathbb{R}^2$. The $\delta-\epsilon$-definition there is that $\lim_{\mathbf{x} \to \mathbf{x_0}} f(\mathbf{x}) = L$ if and only if

$$\vert \mathbf{x} - \mathbf{x_0} \vert < \delta_1 \text{ implies } \vert f(\mathbf{x}) - L\vert < \epsilon.$$

Yet for $\delta_1$ in the role of "$\epsilon$", continuity of $\gamma$ implies that there exists $\delta$ such that

$$\vert t - t_0 \vert < \delta \text{ implies } \vert \mathbf{x} - \mathbf{x_0} \vert < \delta_1$$

Concatenating, $$\vert t - t_0 \vert < \delta \text{ implies } \vert f(\mathbf{x}) - L\vert < \epsilon.$$

In short, continuity of $\gamma$ is what guarantees that that you're in a close-enough neighborhood of $\mathbf{x_0}$ to get whatever general estimate of closeness you get, as in @Git Gud's answer.

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