I'm brushing up on my calculus proofs, and I'm trying to show all the limit laws like $\lim_{x \to c} f(x) + \lim_{x \to c} g(x) = \lim_{x \to c} (f(x) + g(x))$, and similar for subtraction, multiplication, powers, etc.

But I'm sick of finding the exact $\delta$s that work. I know how to prove that if $f$ is continuous at $c$, then $f(\lim_{x \to c} g(x)) = \lim_{x \to c} f(g(x))$. I would like to extend this to the case where $f$ takes two inputs, that way, I can define $h(x,y)$ to be $x + y$, and then the only thing I need to do is show that $h$ is continuous. Then the proof for the sum law would go like this:

$$\lim_{x \to c} f(x) + \lim_{x \to c} g(x) = h(\lim_{x \to c} f(x), \lim_{x \to c} g(x)) = \lim_{x \to c} h(f(x), g(x)) = \lim_{x \to c} (f(x) + g(x))$$

I don't know if the middle equal sign is legitimate though, because there are two limits I'm pulling out instead of one. I can't seem to justify that step, but I suppose part of the trouble is "What does it mean for a function to have two inputs?". (I'm still in the existential, "what is this this object, really?" phase, and I've yet to get a good enough grounding in set/category/type theory to express certain questions).

My questions are:

  • Would I have to show $h$ is continuous when varying just one variable, or actually continuous? (I'm thinking of functions like $z = \frac{x^2 - y^2}{x^2 + y^2}$, where the limit at the origin doesn't actually exist.)
  • Is a step like $\langle \lim_{x \to c} f(x), \lim_{x \to c} g(x) \rangle = \lim_{x \to c} \langle f(x), g(x) \rangle$ legitimate? This way, I can make $h$ accept a vector as input, and just use the one-variable "limits commute with continuous functions" theorem.

You need that $h\colon \mathbb R^2\to\mathbb R$ is continuous. Note that you do esssentially the same looking-for-nice-$\delta$ stuff to show that addition is continuous as you would to show additivity of $\lim$ directly.

The step $\langle \lim f,\lim g\rangle=\lim\langle f,g\rangle$ needs to be justified, but that is straightforward (or in fact more or less the definition of product topology on $\mathbb R^2$).

  • $\begingroup$ Okay, so this approach still runs into the same issues, so I guess I'll just do it directly. If I wanted to prove that limit and vector statement though, would I use different distance functions for the $\mathbb{R}$ and $\mathbb{R^2}$ limits, right? And the statement works because those two functions are somehow compatible? $\endgroup$ Sep 13 '13 at 6:23

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