tag removed - 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 $int_a^bf(x)dx = lim_{hto 0} sum_{n=1}^{(b-a)/h}hcdot f(a+hn)$
2: Let us replace our measly f (x) with its definition, in terms of the derivative of F (x), namely $f(x) = lim_{jto 0}frac{F(x+j)-F(x)}{j}$. Thus, our first equation becomes

$lim_{hto 0} sum_{n=1}^{(b-a)/h}hcdot lim_{jto 0}frac{F(a+hn+j)-F(a+hn)}{j}$

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:
$lim_{hto 0} sum_{n=1}^{(b-a)/h}hcdot frac{F(a+h(n+1))-F(a+hn)}{h}$, and the *h*s can cancel out: $lim_{hto 0} sum_{n=1}^{(b-a)/h}F(a+h(n+1))-F(a+hn)$.
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?

Thanks for your help!
-Gabriel Benamy