Proving the Limit of a Product?

sponsoredwalk

In trying to prove the limit product rule I've found all explanations eventually hit on a point where I lose all understanding.

1: If [latex] \lim_{x \to c} f(x) \ = \ L \ and \ \lim_{x \to c} g(x) \ = \ M \ [/latex]

We define the limit;

[latex] \ \forall \ \epsilon \ >\ 0 \ \exists \ \delta > 0 \ : \ \forall \ x \ ,[/latex]

[latex] \ 0\ < \ | \ x \ - \ c \ | < \delta \ \Rightarrow \ 0 \ < \ | \ f(x)g(x) \ - \ LM \ | \ < \ \epsilon [/latex]



2: Rewrite [latex] f(x) \ = \ L \ + \ (f(x) \ - \ L) \ and \ g(x) \ = \ M \ + \ (g(x) \ - \ M) [/latex]

3: Rewrite [latex] f(x)g(x) \ - \ LM \ as[/latex]

[latex] [L \ + \ (f(x) \ - \ L)] \ [ M \ + \ (g(x) \ - \ M) ] \ - \ LM \ = [/latex]

[latex]LM \ + \ L(g(x) \ - \ M) \ + M(f(x) \ - \ L) \ + \ (f(x) \ - \ L)(g(x) \ - \ M) \ - \ LM [/latex]

[latex] L(g(x) \ - \ M) \ + M(f(x) \ - \ L) \ + \ (f(x) \ - \ L)( g(x) \ - \ M) [/latex]

All this I'm fine with, but in the next oart each source I've read confuses me. I'll give the one from Thomas Calculus.

"Since f & g have limits L & M as x-->c, ∃ positive numbers δ_1, δ_2, δ_3, δ_4 such that ∀ x;

[latex]0 \ < \ |x \ - \ c| \ < \delta_1 \Rightarrow \ |f(x) \ - \ L| \ < \ \sqrt{ \frac{ \epsilon }{3} } [/latex]

[latex] 0 \ < \ |x \ - \ c| \ < \delta_2 \Rightarrow \ |g(x) \ - \ M| \ < \ \sqrt{ \frac{ \epsilon }{3} } [/latex]

[latex] 0 \ < \ |x \ - \ c| \ < \delta_3 \Rightarrow \ |f(x) \ - \ L| \ < \ \frac{ \epsilon }{3(1 \ + \ |M|) } [/latex]

[latex] 0 \ < \ |x \ - \ c| \ < \delta_4 \Rightarrow \ |g(x) \ - \ M| \ < \ \frac{ \epsilon }{3(1 \ + \ |L|) } [/latex]

What does this even mean and where does it come from???

Where do the epsilon parts on the right come from and that crazy thing under the square root, wtf!


Then; the absolute value in the original equation is equivalent to;

[latex] |f(x)g(x) \ - \ LM| \le \ | L(g(x) \ - \ M) \ + M(f(x) \ - \ L) \ + \ (f(x) \ - \ L)( g(x) \ - \ M) | [/latex]

[latex] \le \ | L | \ | (g(x) \ - \ M) | \ + |M| \ | (f(x) \ - \ L) | \ + \ | (f(x) \ - \ L)| |( g(x) \ - \ M) | [/latex]

but then my book goes off writing the following which I have no idea where it came from nor why you'd do it nor how you'd figure out that this is what you do.

[latex] \le ( 1 \ + \ |L| ) |g(x) \ - \ M | \ + \ (1 \ + \ |M| ) |f(x) \ - \ L | \ + \ |f(x) \ - \ L| |g(x) \ - \ M| [/latex]

Then this becomes < [latex] \frac{ \epsilon }{3} \ + \ \frac{ \epsilon }{3} \ +\sqrt{ {\frac{ \epsilon }{3}} } \sqrt{ {\frac{ \epsilon }{3}} [/latex]

And I'm lost

Fremen

Edit: Damn + signs aren't displaying properly. Anyone help?

This stuff can be a little tricky, especially if you haven't seen it before. You can take some consolation in the fact that college maths doesn't actually get a whole lot harder than the epsilon-delta stuff, just more abstract.

There's a certain knack to reading proofs that takes a little while to pick up. On a first reading, you don't necessarily want to follow the whole line of the argument, just convince yourself that everything is true, and that one line follows from the next.

sponsoredwalk wrote: »

In trying to prove the limit product rule I've found all explanations eventually hit on a point where I lose all understanding.

"Since f & g have limits L & M as x-->c, ∃ positive numbers δ_1, δ_2, δ_3, δ_4 such that ∀ x;

[latex]0 \ < \ |x \ - \ c| \ < \delta_1 \Rightarrow \ |f(x) \ - \ L| \ < \ \sqrt{ \frac{ \epsilon }{3} } [/latex]

[latex] 0 \ < \ |x \ - \ c| \ < \delta_2 \Rightarrow \ |g(x) \ - \ M| \ < \ \sqrt{ \frac{ \epsilon }{3} } [/latex]

[latex] 0 \ < \ |x \ - \ c| \ < \delta_3 \Rightarrow \ |f(x) \ - \ L| \ < \ \sqrt{ \frac{ \epsilon }{3(1 \ + \ |M|} } [/latex]

[latex] 0 \ < \ |x \ - \ c| \ < \delta_4 \Rightarrow \ |g(x) \ - \ M| \ < \ \sqrt{ \frac{ \epsilon }{3(1 \ + \ |L|} } [/latex]

Ok, let's forget about why he's doing it. Is it true?
Well, set the crazy crap on the right hand side equal to a load of new variables, say [latex] \epsilon_1, \dots, \epsilon_4 [/latex]
Then from our assumptions about the behaviour of f and g, we can see that it follows (from the definition of a limit).

Then; the absolute value in the original equation is equivalent to;

[latex] |f(x)g(x) \ - \ LM| \le \ | L(g(x) \ - \ M) \ + M(f(x) \ - \ L) \ + \ (f(x) \ - \ L)( g(x) \ - \ M) | [/latex]

[latex] \le \ | L | \ | (g(x) \ - \ M) | \ + |M| \ | (f(x) \ - \ L) | \ + \ | (f(x) \ - \ L)| |( g(x) \ - \ M) | [/latex]

but then my book goes off writing the following which I have no idea where it came from nor why you'd do it nor how you'd figure out that this is what you do.

I trust you're happy that this part is true then? It's just the triangle inequality. With regard to figuring out that this is what you do, that's a tricky one. This probably isn't the first version of the proof that there ever was. In all likelihood, someone worked through tens or hundreds of special cases until it became second nature, then generalised the method.

[latex] \le ( 1 \ + \ |L| ) |g(x) \ - \ M | \ + \ (1 \ + \ |M| ) |f(x) \ - \ L | \ + \ |f(x) \ - \ L| |g(x) \ - \ M| [/latex]

Another useful trick: look at the previous line, look at the current line, and try to work out what changed. Forget about everything else (temporarily).
Here, he's added
[latex] |g(x) \ - \ M | \ [/latex] and [latex] |f(x) \ - \ L | \ [/latex]
to the equation, and re-arranged. The inequality still holds since he's added something non-negative to the right.

Now, we use our expressions involving [latex] \epsilon_1, \dots, \epsilon_4 [/latex] and [latex] \delta_1, \dots, \delta_4 [/latex]

Then this becomes < [latex] \frac{ \epsilon }{3} \ + \ \frac{ \epsilon }{3} \ +\sqrt{ {\frac{ \epsilon }{3}} } \sqrt{ {\frac{ \epsilon }{3}} [/latex]

So now we see why we needed the square roots: it's because, via an application of the triangle inequality, we need to multiply [latex]|f(x) \ - \ L|[/latex] and [latex] |g(x) \ - \ M| [/latex] together.


Are sure you meant to put square roots in the [latex]\delta_3[/latex] and [latex]\delta_4[/latex] inequalities? All we need to do is multiply them by [latex]1+ |M|[/latex] and [latex]1 \ + \ |L|[/latex], so it would make sense if there was no root there.

sponsoredwalk

I can tell you that the single most painful thing I am encountering with my studies of math is sifting through the equations that come prepackaged, like that crazy epsilon fraction we see here, to try to build everything from first principles.

So, I understand that we just did a lot of work to arrive at;

[latex] |f(x)g(x) \ - \ LM| \le \ | L | \ | (g(x) \ - \ M) | \ + |M| \ | (f(x) \ - \ L) | \ + \ | (f(x) \ - \ L)| |( g(x) \ - \ M) | [/latex]

and by playing around with the R.H.S. it will prove the L.H.S. It's the same thing you do when proving the sum of limits. The thing is, with the sum of limits it's all so intuitive I could rattle it off now without memorizing some crazy fraction that leaps out of nowhere. I'm so depressed to read that even Apostol uses this bloody fraction to prove the equation and there has got to be a better way.

BTW: Yeah the fraction in δ_3 & δ_4 was a mistake, It's fixed now! Also, the + signs tend to do that for me too sometimes but the just reappear

All I notice is that |L| magically became (1 + |L|), |M| magically became (1 + |M|) and that we are magically adding on a (1 + |L|) & (1 + |M|) to a magically squared out √(ε/3) in the numerator

I also don't unerstand why you need 4 delta's tbh

Fremen

Try going back to the principle of "what information do you have, and how can you use it to prove what you want?".

In this case, you have information about f and g individually, but you want to prove something about (f)(g). If you were trying to do this on your own, for the first time, you might start at step three.
Here, they've used a smart way of decomposing (f)(g)-LM in terms of stuff you already know about.

Once you do that, you just need to find use bounds on the stuff you already know about. If you can show that everything in your decomposition can be made to sum to something less than epsilon, you're done.

Since there are three terms, you might as well try to show that each term can be made smaller than epsilon/3. One hard problem has been broken up into three easy ones.

It's only THEN that you'd start thinking about crazy bounds like your delta1,2,3 and 4. In a sense the solution in the book has been presented backwards.

Edit: forget about the 1+M bit for now. The only reason they go to the trouble to move from M to 1+M is to make sure that everything still makes sense when M=0.

bnt

Fremen wrote: »

Edit: Damn + signs aren't displaying properly. Anyone help?

I noticed this before, it seems to have something to do with Quick Edit mode. It doesn't happen if you select "Go Advanced" when editing a post containing LaTeX.

sponsoredwalk

Thanks for the help fremen, I think I've got it now