There's a way to see how to do this, but I'm gonna make the problem a bit more abstract first... Let's say the problem was
"Find the limit of ∫ f(x) dx from a to a+h all over h as h approaches 0." (a is a number, in your case it's 1)
Then, ∫ f(x) dx = F(x) + C (where F(x) is an antiderivative of f(x), and the integral of f(x) from x=a to x=a+h is given by F(a+h) - F(a). (1st Fundamental Theorem of Calculus)
Thus, since you have to divide this integral by h, you have to find the limit of:
lim h→0 (F(a+h) - F(a)) / h .
Here's where it gets cool. You probably recognize that that limit is equal to F'(a), a.k.a. f(a).
In the case of your problem, a=1 so all you have to do is plug in x = 1 into sqrt(x^5+8)! That's where the 3 came from haha.
I also wanna note that Steve's idea with the rectangles is good: the integral value *does* approach 0. As the rectangle closes in on x = 1, its height is 3 since that's the y-value you get when you plug in x = 1 into sqrt(x^5+8). The width of the rectangle is h. So the integral's value tends to 3h. But the problem asks you to divide by h, which is why the limit approaches 3.
I feel like my presentation was kinda confusing so please down-thumb me if you agree! Thx :)