How to simplify my arc length formula. When we use Riemann Sums to evaluate definite integrals, and tend the limit of width of the rectangle to zero, then the area become zero. 
If I use similar logic to evaluate arc length (not the traditional formula), by- 
$l^2=h^2+(f(x+h)-f(x))^2$, where $l$ is arc length, and do $$\lim\limits_{h\to0} \sqrt{h^2+(f(x+h)-f(x))^2}+\cdots+\sqrt{h^2+(f(x+nh)-f(x+(n-1)h))^2}$$ What will it be?
 A: We need to get our head around the fact that the number of summands is becoming arbitrarily large, even as individual summands are approaching zero. To put it in stark terms, consider the limit
$$\lim_{n\to\infty} \sum_{j=1}^n \frac 1n$$
By the 'naive' argument, "Look, each term $\frac 1n$ becomes zero in some sense, so isn't the limit of the sum zero?".
No. Because the number of summands offsets the smallness of the individual summands. And in this case, that sum is equal to one for all $n \geq 1$ and hence so is the limit.
By the way, the sum above is one Riemann sum expression for the integral
$$\int_0^1 1 \ dx$$
which of course is equal to one.

Added:
Rewriting your formula, the arc length of a smooth function $f$ over an interval $[a,b]$, with a step size $h = \frac{b-a}{n}$ is the limit
$$\lim_{n\to\infty} \sum_{j=1}^n \sqrt{h^2 + \left(f(a + jh) - f(a + (j-1)h) \right)^2}$$
This is equal to
$$\lim_{n\to\infty} \sum_{j=1}^n \sqrt{1 + \left(\frac{f(a + jh) - f(a + (j-1)h)}{h}\right)^2} \cdot h $$
In the limit as $n\to\infty$ or $h\to 0$, the inner quotient is the derivative and we recover the traditional formula.
