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.

It is not actually rigorous what he says... but what he means is that, when $\Delta x $ approches $0$, then $f$ is approximately constant (assuming it is continuous). What you can do is, for instance:
Fix $\epsilon >0$. There exists $\delta>0$ such that: $f(x)-\epsilon<f(u)< f(x)+\epsilon$ for all $u$ in the $\delta$-neighbourhood of $x$. Therefore:
$ (f(x)-\epsilon) \Delta x \leq \int_x^{x+\Delta x} (f(x)- \epsilon) du \leq \int_x^{x+\Delta x} f(u)du \leq \int_x^{x+\Delta x} (f(x)+ \epsilon) du =(f(x)+\epsilon) \Delta x $
which implies:
$$\displaystyle f(x)-\epsilon \leq \lim_{\Delta x \rightarrow 0}\frac{\int_x^{x+\Delta x} f(u)du}{\Delta x}\leq f(x)+\epsilon$$
Since this holds for every $\epsilon>0$, we have that 
$$\displaystyle \lim_{\Delta x \rightarrow 0}\frac{\int_x^{x+\Delta x} f(u)du}{\Delta x}=f(x)$$
