Why $\lim_{\Delta x\to 0} \cfrac{\int_{x}^{x+\Delta x}f(u) du}{\Delta x}=\cfrac{f(x)\Delta x}{\Delta x}$? I'm reading Nahin's: Inside Interesting Integrals.



I've been able to follow it until: 
$$\lim_{\Delta x\to 0} \cfrac{\int_{x}^{x+\Delta x}f(u) du}{\Delta x}=\cfrac{f(x)\Delta x}{\Delta x}$$
I don't know what justifies the passage of the limit here. I've tried to write it as:
$$\lim_{\Delta x\to 0} \left[ \cfrac{1}{\Delta x}\cdot \int_{x}^{x+\Delta x}f(u) du\right]=\left[\lim_{\Delta x\to 0} \cfrac{1}{\Delta x}\right]\cdot \left[ \lim_{\Delta x\to 0} \int_{x}^{x+\Delta x}f(u) du \right]$$
But it seemed to create more non-sense than I previously had.

Let us consider $$A= \cfrac{\int_{x}^{x+\Delta x}f(u) du}{\Delta x} $$ and let $F(u)=\int f(u)du$. So the numerator is $F(x+\Delta x)-F(x)$ which is $\Delta x \,F'(x)=\Delta x \,f(x)$.
