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.
 A: If $f$ is continuous on $[x,x+\Delta x]$, then by the MEAN VALUE THEOREM there exists a number $\xi\in [x,x+\Delta x]$ such that
$$\int_x^{x+\Delta x} f(x)\,dx=f(\xi)\Delta x$$
and as $f$ is continuous we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{\Delta x\to 0}\frac{1}{\Delta x}\int_x^{x+\Delta x} f(x)\,dx=\lim_{\Delta x\to 0}f(\xi)=f(x)}$$
A: 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 
$$\displaystyle \lim_{\Delta x \rightarrow 0}\frac{\int_x^{x+\Delta x} f(u)du}{\Delta x}=f(x)$$
A: 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.
A: 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 $\Delta x \,F'(x)=\Delta x \,f(x)$.
