You have a slight problem. You must require $t_1 ≠ t_2$ in your double-limit, otherwise it is ill-defined. That said, with that error fixed, it's a fine way to think of "instantaneous rate of change", though I don't quite believe that it is an intuitive way. In fact, I never had the idea that it was a double-limit until after I read your post! Haha.. — user21820 10 hours ago

Uhh why didn't you just use "$t_1≠t_2$" as I said in my first comment? Your second version is still wrong because it doesn't exclude a singularity at $t_a$ and so it is not equivalent to the one dimensional limit... — user21820 2 hours ago

@user21820: ? $t_1 < t_a < t_2$ implies $t_1 \ne t_2$. Note that those are strict inequalities, so by trichotomy, $t_1 \ne t_2$. I'm also not sure where you're imagining "a singularity at $t_a$". In $f$? In that case, $f$ is not differentiable at $t_a$. In the quotient? There is always a removable singularity whenever $f$ is differentiable, but if there's a non-removable one, that can only come from $f$ itself. Counterexample? — The_Sympathizer 1 hour ago

Also $t_1 \ne t_2$ is not strong enough. Suppose that $t_1$ and $t_2$ approach each other while also approaching $t_a$, but they do not bound $t_a$ between them. If they approach each other "faster" than they approach $t_a$, and $f$ is suitably ill-behaved, the IRoC might fail to exist even while the derivative, defined with the 1-D limit, still does exist. — The_Sympathizer 1 hour ago

A pathological example is provided by $f(x) := \begin{cases} x^2,\ x \in \mathbb{Q} \\ -x^2,\ x \notin \mathbb{Q}\end{cases}$, which is differentiable at 0 but nowhere else. Consider the sequence of $a_j$ and $b_j$ given by $b_j := \frac{2}{2^j}$ and $a_j := -\frac{\pi}{2^j}$, so that $b_j$ approaches $0$ along rational numbers and $a_j$ approaches along irrational numbers. Now take $t_{1_j} := a_j - \frac{10}{j}$ and $t_{2_j} := b_j - \frac{10}{j}$. We now have $\lim_{j \rightarrow \infty} t_{1_j} = \lim_{j \rightarrow \infty} t_{2_j} = 0$, but the IRoC as defined blows up to infinity. — The_Sympathizer 1 hour ago

You're wrong. Please think carefully. If the double-limit I stated exists, then it obviously implies the standard limit. In contrast, your second version doesn't exclude a singularity as I stated. Furthermore, it is silly to say that my version fails if you choose $t_1,t_2$ to follow a certain path, because your version is even worse. No, double-limits are over all paths. And it is a basic real analysis fact that the standard limit implies continuity of $f$ at $t_a$, whereas your second version does not. — user21820 1 hour ago

@user21820: You have not given the express counterexample featuring the kind of singularity you're talking about that is not excluded by the $t_1 < t_a < t_2$ condition, nor shown what was wrong with my counterexample. Also, there is no "worse" - either the limit exists or it doesn't, period. One bad path is sufficient, so the limit does not exist enough times under only the hypothesis $t_1 \ne t_2$ as my example proves. It's not strong enough. Unless you can post a problem with that example, which you have not done, simply asserted a problem. You have not shown a problem. — The_Sympathizer 8 mins ago

On the other hand, I gave an explicit example that shows both the original formulation I gave was wrong and $t_1 \ne t_2$ is insufficient. An alternative might be to take $\lim_{(\epsilon_1, \epsilon_2) \rightarrow (0, 0)} \frac{f(t_a + |\epsilon_2|) - f(t_a - |\epsilon_1|)}{|\epsilon_2| - |\epsilon_1|}$ which forces the necessary $t_1 < t_a < t_2$ condition without requiring a limitation on paths and hence is the full standard "two dimension limit" condition. — The_Sympathizer 6 mins ago

Look, I have no idea why you don't get it. Please look up the rigorous definition of the double-limit, and prove for yourself that my version works (it's a standard textbook exercise). In contrast, your version fails whenever $f$ has a removable discontinuity at $t_a$ (whereas mine doesn't). And your third version "$\lim_{(\epsilon_1, \epsilon_2) \rightarrow (0, 0)} \frac{f(t_a + |\epsilon_2|) - f(t_a - |\epsilon_1|)}{|\epsilon_2| - |\epsilon_1|}$" in your last comment is completely bogus. — user21820 4 mins ago