# Limit Proof

1. Apr 25, 2010

In trying to prove the limit product rule I've found all explanations

to hit on a point where I lose understanding.

1: If $$\lim_{x \to c} f(x) \ = \ L \ and \ \lim_{x \to c} g(x) \ = \ M \$$

We define the limit as;

$$\ \forall \ \epsilon \ >\ 0 \ \exists \ \delta > 0 \ : \ \forall \ x \ \rightarrow \ 0\ < \ | \ x \ - \ c \ | < \delta \ \Rightarrow \ 0 \ < \ | \ f(x)g(x) \ - \ LM \ | \ < \ \epsilon$$

2: Rewrite $$f(x) \ = \ L \ + \ (f(x) \ - \ L) \ and \ g(x) \ = \ M \ + \ (g(x) \ - \ M)$$

3: Rewrite $$f(x)g(x) \ - \ LM \ as$$

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

$$LM \ + \ L(g(x) \ - \ M) \ + M(f(x) \ - \ L) \ + \ (f(x) \ - \ L)(g(x) \ - \ M) \ - \ lm$$

$$L(g(x) \ - \ M) \ + M(f(x) \ - \ L) \ + \ (f(x) \ - \ L)( g(x) \ - \ M)$$

All this I'm fine with, but next 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;

$$0 \ < \ |x \ - \ c| \ < \delta_1 \Rightarrow \ |f(x) \ - \ L| \ < \ \sqrt{ \frac{ \epsilon }{3} }$$

$$0 \ < \ |x \ - \ c| \ < \delta_2 \Rightarrow \ |g(x) \ - \ M| \ < \ \sqrt{ \frac{ \epsilon }{3} }$$

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

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

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

2. Apr 26, 2010

### CompuChip

How you find these numbers, is perhaps a bit unclear to you. However, if you follow the proof through to the end, you will hopefully see what it does.

So apart from all this magic with the epsilons and delta's, do you agree with the statement? When you look at the definition for limit, you probably will. For example, since f(x) has limit L, I can always make |f(x) - L| as small as I want. In symbols,
$$\ \forall \ \epsilon' \ >\ 0 \ \exists \ \delta' > 0 \ : \ \forall \ x \ \rightarrow \ 0\ < \ | \ x \ - \ c \ | < \delta' \ \Rightarrow \ 0 \ < \ | \ f(x) \ - \ L \ | \ < \ \epsilon'$$ (*)
Now what the proof does, is simply pick two such $\epsilon'$ (namely $\sqrt{\epsilon / 3}$ and $\sqrt{\epsilon / 3(1 + |M|)}$ where $\epsilon, M$ are given numbers, then the existence of the limit ensures that I can find values of delta' for which (*) is true.

The proof then probably goes on to pick the smallest delta of the four, such that all four estimates hold simultaneously.
Then you can plug all those estimates into
$$f(x) g(x) - L M = L(g(x) \ - \ M) \ + M(f(x) \ - \ L) \ + \ (f(x) \ - \ L)( g(x) \ - \ M)$$
and show that it is smaller than $\epsilon$.

3. Apr 26, 2010

Hey thanks for the reply, well I did try to follow the proof forwards but not only does that crazy value for epsilon scare the **** out of me but I just get confused. I'l show you where;

I should have written a bit more; the absolute value in the original equation is equivalent to;

$$|f(x)g(x) \ - \ LM| \le \ | L(g(x) \ - \ M) \ + M(f(x) \ - \ L) \ + \ (f(x) \ - \ L)( g(x) \ - \ M) |$$

$$\le \ | L | \ | (g(x) \ - \ M) | \ + |M| \ | (f(x) \ - \ L) | \ + \ | (f(x) \ - \ L)| |( g(x) \ - \ M) |$$

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.

$$\le ( 1 \ + \ |L| ) |g(x) \ - \ M | \ + \ (1 \ + \ |M| ) |f(x) \ - \ L | \ + \ |f(x) \ - \ L| |g(x) \ - \ M|$$

Then this becomes < $$\frac{ \epsilon }{3} \ + \ \frac{ \epsilon }{3} \ +\sqrt{ {\frac{ \epsilon }{3}} } \sqrt{ {\frac{ \epsilon }{3}}$$

And I'm lost