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's assume $f(x)$ is continuous. Then for every $\Delta x>0$ there exist $x_{\min},x_{\max}\in[x,x+\Delta x]$ such that 
$$
f_{\min}\leq f(t)\leq f(x_\max)
$$
for $t\in[x,x+\Delta x]$. Of course,
\begin{align*}
\lim_{\Delta\to0}f(x_\min)&=f(x)&
\lim_{\Delta\to0}f(x_\max)&=f(x)
\end{align*}
Now, 
$$
f(x_\min)\leq
\frac{1}{\Delta x}\int_{x}^{x+\Delta x}f(t)\,dt\leq
f(x_\max)\tag{1}
$$
Taking $\Delta x\to 0$ in (1) gives the result.
