# Can a limit to zero of a limit to zero assume they're both going at the same rate?

My title can be a bit confusing, so here's a bit of background.
The corollary to the Fundamental Theorem of Calculus says that $\int_a^bf(x)dx = F(b)-F(a)$, assuming that F'(x) = f (x), or that the area under the curve f (x) from x = a to x = b is equal to the difference of values of the antiderivative of f (x) at a and b.

The following is my attempt to prove it.
1: The area under the curve of f (x) from x = a to x = b is equal to the area of the rectangles under the curve as you take more and more rectangles. See this image:
Mathematically speaking, it's $\int_a^bf(x)dx = \lim_{h\to 0} \sum_{n=1}^{(b-a)/h}h\cdot f(a+hn)$
2: Let us replace our measly f (x) with its definition, in terms of the derivative of F (x), namely $f(x) = \lim_{j\to 0}\frac{F(x+j)-F(x)}{j}$. Thus, our first equation becomes

$\lim_{h\to 0} \sum_{n=1}^{(b-a)/h}h\cdot \lim_{j\to 0}\frac{F(a+hn+j)-F(a+hn)}{j}$

Now, my question is, since both h and j are going to zero via a limit, can I assume that they are effectively the same? Can I simply replace all instances of j with an h and rid myself of an unnecessary second limit? If I could, my proof would continue as follows:

3: Replacing all j's with h's yields:
$\lim_{h\to 0} \sum_{n=1}^{(b-a)/h}h\cdot \frac{F(a+h(n+1))-F(a+hn)}{h}$, and the *h*s can cancel out: $\lim_{h\to 0} \sum_{n=1}^{(b-a)/h}F(a+h(n+1))-F(a+hn)$.
4: Thankfully, this becomes a telescoping series, as seen here:
$F(a+h(1))-F(a+0h) + F(a+h(2))-F(a+1h) + F(a+h(3))-F(a+2h) + ... = -F(a-h) + F(b-h)$
$+ F(a+h(\frac{b-a}{h}))-F(a+h(\frac{b-a}{h}-1) = F(b) - F(b-h)$
which, together, yields -F (a - h) + F (b) as the sum.
Putting this back in, we get $\lim_{h \to 0} -F(a - h) + F(b) = F(b) - F(a) = \int_a^bf(x)dx = F(b)-F(a)$

However, steps 3 and 4 require the ability for me to assume that h and j are the same thing. My teacher (who admittedly doesn't deal with this too often), whom I asked first, said that perhaps h and j are going to 0 at different rates. However, I do not think that the concept of a limit to 0 at a rate actually means anything.
So the question I bring to you is: Is the operation that I performed to go from step 2 to step 3 a valid operation? If so, why? If not, why not?

-Gabriel Benamy

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"However, I do not think that the concept of a limit to 0 at a rate actually means anything." It means rather a lot. When you send two things to zero, the value of the resulting expression depends heavily on the relative rate at which they go to zero, as Douglas demonstrates. The definition of a multivariable limit is more restrictive than the definition of two single-variable limits. – Qiaochu Yuan Feb 23 '10 at 21:13
*in general. In the nicest cases, it doesn't matter, but you can't assume you are in those cases until you prove it. – Qiaochu Yuan Feb 23 '10 at 21:14
So in this particular case, how would I go about finding the relative rates at which these two variables are going to 0? I haven't taken multivariable calculus yet, so I wouldn't know. Thanks, though. My teacher will be pleased to know that he was on the right track. – Gabriel Benamy Feb 23 '10 at 21:22
@Gabriel: that question is meaningless. In your double limit, you first take the limit of the expression as j tends to zero. That gives you something that still depends on h, and you take the limit of that as h tends to zero. There are no "rates." – Darsh Ranjan Feb 24 '10 at 4:32

In general, no. You need some continuity assumption.

$$\lim_{x\to 0} \lim_{y\to 0} \frac yx = \lim_{x\to 0} 0 = 0$$ $$\lim_{x\to 0} \frac xx = 1$$

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I wish to respectfully point out a suboptimality in your approach: you are trying to prove the Fundamental Theorem of Calculus by yourself and by talking to a (possibly not very knowledgeable, but I don't know the whole story) teacher. This is not necessary: there are literally thousands of books which contain complete, detailed proofs: e.g. any calculus textbook. (It seems that most calculus classes at the high school and college freshman level do not go into a careful discussion of the proof of the fundamental theorem of calculus, but nevertheless the textbooks still cover it, for edification of the most inquisitive/intelligent 1% of student readers and to make the math professors happy -- or rather, less unhappy; see below! -- with their texts.)

On the other hand, if you look more deeply into things, you may find that there are some unexpected subtleties. For a careful statement of FTC, see e.g.

http://math.uga.edu/~pete/243integrals1.pdf

and especially Theorem 11 on page 8. (These are lecture notes for a course on undergraduate real analysis, i.e., one or two levels beyond freshman calculus.) In particular one needs to pay some attention to what the Riemann integral actually means: it's not just equally spaced subintervals and evaluation at the endpoints; it's more general than that. This has both a positive and negative consequence:

Good news: Assuming that f is Riemann integrable, you can use the flexibility of the choice of the sample point in the subinterval to get a very quick proof of the part of FTC you're asking about, as an application of the Mean Value Theorem.

Bad news: You need to assume that f is Riemann integrable for this version of FTC to be true. Strictly speaking this is necesary in order to define the left hand side, but the surprising part is that assuming that f has an antiderivative F is not in general enough to ensure Riemann integrability. (There are fancier versions of the integral which overcome this.)

It is a fact that every continuous function on [a,b] is Riemann integrable, and in calculus this is usually a mild enough hypothesis. However, the proof of the fact that every continuous function is Riemann integrable is much harder than the proof of the rest of FTC! (This is the jumping off point of the notes I cited.) This point is shoved under the rug in every mainstream (i.e., non-Spivak, etc.) freshman calculus text I have ever seen, to my annoyance.

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Unrelated question: Isn't FTC a corollary in spivak? – Harry Gindi Feb 24 '10 at 6:56
Seconded. A more succinct way to put this is that a standard high school or introductory college course in calculus does not prepare you for the actual proof of the FTC. (It usually doesn't even prepare you to understand the actual definition of the Riemann integral!) – Qiaochu Yuan Feb 24 '10 at 6:56
@QY: What are you saying, that I could be more succinct? I have a 17 page rebuttal to this on my webpage; would you like me to send you the link? – Pete L. Clark Feb 24 '10 at 7:00
I've never seen a proof that uses the Riemann integrability criterion directly. All of the proofs I've seen first prove that riemann integrability is equivalent to darboux integrability. Then they just prove things about darboux integrability. – Harry Gindi Feb 24 '10 at 7:05
@GB: It's not so much that there are plenty of proofs of FTC; rather the same proof is written down in many different sources. There are lots of opportunities for being creative and working out things for yourself mathematics. In my opinion, this is not such a good one: your time would be better spent learning the usual proof, which is more subtle than you seem to realize. About other applications of your technique: no, it is unfortunately not valid, despite the fact that it gives correct answers much of the time. – Pete L. Clark Feb 24 '10 at 23:51

Your proof is wrong as stated, but it's pretty easy to rescue it. In your sum $\sum_{n=1}^{(b-a)/h}h\cdot f(a+hn)$, you actually don't have to evaluate $f$ at $a+hn$; you can evaluate it at any point in the interval $[a+h(n-1), a+hn]$ and the sum will still converge to the integral in the limit as $h$ tends to zero. That's obvious pictorially (if you think of the integral as an area). Now, by the mean-value theorem (which is also obvious pictorially), there is a point $x^*_n$ in $[a+h(n-1), a+hn]$ such that $f(x^*_n) = F'(x^*_n)=(F(a+hn) - F(a + h(n-1)))/h$. If we choose $x^*_n$ as our point of evaluation, then we have $$\sum_{n=1}^{(b-a)/h}h\cdot f(x^*_n) = \sum_{n=1}^{(b-a)/h}(F(a+hn) - F(a + h(n-1))).$$Voilà, telescoping sum.

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