0
$\begingroup$

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.
$\endgroup$
1
$\begingroup$

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 $\mathbb R^2$).

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.