# Thread: Missing Step in Apostol Proof of Basic Limit Theorems?

1. ## Missing Step in Apostol Proof of Basic Limit Theorems?

Hi Folks: in Apostols's Calculus, section 3.5, Proofs of Basic Limit Theorems, page 136, Proofs of (i) and (ii), he makes an assumption which I'm hoping someone can explain to me how he justifies.

He's proving the first limit theorem, that the limit of a sum is the sum of the limits. He says "Since the two statements lim f(x) = A and lim[f(x)-A]=0 are equivalent..."

I can readily see that lim f(x) - A = 0 but don't you need the second limit theorem, that the limit of a difference is the difference of the limits, to therefore conclude that lim[f(x)-A]=0?

2. ## Re: Missing Step in Apostol Proof of Basic Limit Theorems?

Originally Posted by MichaelLitzky
in Apostols's Calculus, section 3.5, Proofs of Basic Limit Theorems, page 136, Proofs of (i) and (ii), he makes an assumption which I'm hoping someone can explain to me how he justifies.
He's proving the first limit theorem, that the limit of a sum is the sum of the limits. He says "Since the two statements lim f(x) = A and lim[f(x)-A]=0 are equivalent..."
I can readily see that lim f(x) - A = 0 but don't you need the second limit theorem, that the limit of a difference is the difference of the limits, to therefore conclude that lim[f(x)-A]=0?
I think that only someone with with a copy of that particular textbook can really answer your question. I no longer have a copy of Tom Apostol's book and don't really remember how he treats limits. But in general, that follows at once from a standard $\epsilon(\delta)$ definition.

Informally: if $f(x) \approx A$ 'near' then $f(x)-A \approx 0$. Because by $B \approx A$ means that $B-A$ can be made small as needed.

But hopefully someone else has a copy of that text.

3. ## Re: Missing Step in Apostol Proof of Basic Limit Theorems?

Thanks, Plato. I think I can rephrase the question so it doesn't directly reference the Apostol text.

So yes, by the $\epsilon$/ $\delta$ definition of a limit we know that $\lim_{x\to p}f(x)=A$ means |f(x)-A|< $\epsilon$ whenever 0<|x-p|< $\delta$. We now set out to prove the Sum Law ( $\lim_{x\to p}[f(x)+g(x)]=\lim_{x\to p}f(x)+\lim_{x\to p}g(x)$). We have not yet proven the Difference Law ( $\lim_{x\to p}[f(x)-g(x)]=\lim_{x\to p}f(x)-\lim_{x\to p}g(x)$).

Our first step is to assume that $\lim_{x\to p}f(x)=A$ and $\lim_{x\to p}[f(x)-A]=0$ are the same, which is what Apostol does. Now of course if we already have the Difference Law, then:
$\\ \lim_{x\to p}[f(x)-A]=\\ \lim_{x\to p}f(x)-\lim_{x\to p}A=\\A-A=0$

My question is, without having proven the Difference Law, how can we justify the assumption that $\lim_{x\to p}[f(x)-A]=0$? (I'm trying to get better at proofs, which is why I'm being so nit-picky about when you can assume what.) Thanks!

4. ## Re: Missing Step in Apostol Proof of Basic Limit Theorems?

I think what is being glossed over here, is that A is just a constant, it has no dependence on x. There is a world of difference between asserting:

1) $\lim_{x \to p} [f(x) - g(x)] = \lim_{x \to p}f(x) - \lim_{x \to p}g(x)$

and

2) $\lim_{x \to p} [f(x) - C] = \lim_{x \to p}f(x) - C$

Let's prove #2:

for convenience, we will take $\lim_{x \to p} f(x) = A$, in other words, we are assuming f HAS a limit A at the point p. Of course what this means is that for every $\epsilon > 0$ there does indeed exist some $\delta > 0$ so that whenever:

$0 < |x - p| < \delta$, we have: $|f(x) - A| < \epsilon$.

What we want to show is that for the same $\epsilon$ (which is, after all, arbitrary) we can find a $\delta_1$ so that:

$0 < |x - p| < \delta_1 \implies |(f(x) - C) - (A - C)| < \epsilon$.

Stated THIS way, it is obvious (without invoking "the difference law") because in the last inequality the C's cancel, meaning we can use the same $\delta$!

Taking now C to actually be the limit A, we obtain what Apostol assumes (A - A is obviously 0).

5. ## Re: Missing Step in Apostol Proof of Basic Limit Theorems?

Thank you, Deveno. That makes it crystal clear.