Limit of an integral as $x\to 0$ 
Given an integral function $\displaystyle F(x) = \int_{0}^{x}f(t)dt$ with $f(0)=3$ and $f$ continuous, determine the limit:
$$\lim_{x\to 0}\frac{\displaystyle \int_{0}^{x}f(t)dt}{x}.$$

I know I could apply L'Hôpital's rule or the mean value theorem, but can I state simply that
since $F'(x)=f(x)$ and then $F'(0)=3$, so $$\displaystyle F(x) \;=\; \int_{0}^{x}f(t)dt \;=\; 0+3x+o(x),$$ as $x\to 0$, and so
$$\lim_{x\to 0}\frac{\displaystyle \int_{0}^{x}f(t)dt}{x^3} \;=\; \lim_{x\to 0}\frac{2x}{x} \;=\; 2\;?$$
 A: Your approach is correct provided $f$ is continuous in some small interval $(0,\epsilon)$. However, we can construct some pathological examples where $f$ is discontinuous at countably infinitely many points in the neighborhood of $0$, yet is still Riemann integrable and satisfies the given property. For instance take
$$f(x)=3+\sum_{n=1}^\infty \frac{(-1)^n}{n}\mathbf{1}\left(\left[\frac{1}{n},\frac{1}{n-1}\right]~;~x\right)$$
One can see that, letting $\ulcorner x\urcorner=1/(n_*(x))$ be the smallest unit fraction greater than or equal to $x$ (i.e, $\ulcorner 0.3\urcorner=1/3$ and $n_*(0.3)=3$) that
$$f(x)=3+\sum_{n=n_*(x)-1}^\infty \frac{(-1)^n}{n}\mathbf{1}\left(\left[\frac{1}{n},\frac{1}{n-1}\right]~;~x\right) \\ \leq 3+\sum_{n=n_*(x)-1}^\infty\frac{(-1)^n}{n} $$
And so, using the alternating series,
$$3-\frac{1}{n_*(x)-1}\leq f(x)\leq 3+\frac{1}{n_*(x)-1}$$
Since $\ulcorner x\urcorner=1/(n_*(x))\geq x$ then
$$\frac{1}{n_*(x)-1}>x$$
So
$$3-x\leq f(x)\leq 3+x$$
So finally,
$$\frac{1}{\epsilon}\int_0^\epsilon 3-x ~\mathrm dx\leq\frac{1}{\epsilon}\int_0^\epsilon f(x)\mathrm dx\leq \frac{1}{\epsilon}\int_0^\epsilon 3+x~\mathrm dx \\ 3-\frac{\epsilon}{2}\leq \frac{1}{\epsilon}\int_0^\epsilon f(x)\mathrm dx\leq 3+\frac{\epsilon}{2}$$
Taking the limit as $\epsilon\to 0$ and applying the squeeze theorem gives
$$\lim_{x\to 0}\frac{1}{x}\int_0^xf(t)\mathrm dt=3$$
As required. A very rough idea of the proof is that in the neighborhood of $0$, $f(x)\approx 3$ and so $\int_0^\epsilon f(x)\mathrm dx\approx 3\epsilon$. The above proof makes this more rigorous.
