real analysis - $f$ continous at $x_0$ $⇒lim_{h→0}∫_{x_0}^{x_0+h}frac{f(t)}{h}=f(x_0)$

The function $f:ℝ→ℝ$ is continuous on $x_0\inℝ$.
Prove using the definition of a Darboux Integral that
$$\lim_{h→0}∫_{x_0}^{x_0+h}\frac{f(t)}{h}=f(x_0)$$

I'm a first grade math student following an analysis course. The book that is used is Elementary Analysis by Ross.

Definitions

1. The upper Darboux sum $U(f,P)$ of $f$ with respect to a partition $P$ is the sum
   $$U(f,P)=∑_{k=1}^nM(f,[t_{k-1},t_k])(t_k-t_{k-1})$$




2. The lower Darboux sum $L(f,P)$ of $f$ with respect to a partition $P$ is the sum
   $$L(f,P)=∑_{k=1}^nm(f,[t_{k-1},t_k])(t_k-t_{k-1})$$



3. $f$ is continuous at $x_0$ ⇔ $∀ε>0,∃δ>0,(|x-x_0|<δ ⇒ |f(x)-f(x_0)|<ε)$


4. Let $f$ be a function defined on on $J-\{a\}$ for some interval $J$ containin $a$, and let $L$ be a real number. Then $ \lim_{x→a}f(x)=L$ if and only if
   
   $$∀ε>0,∃δ>0,(0<|x-a|<δ⇒|f(x)-L|<ε)$$

Can someone check if this is an correct proof ?

Proof

Let $ε>0$. Then there exist an $δ>0$, such that: $|x-x_0|<δ ⇒ |f(x)-f(x_0)|<ε$.
Let $0<|h-0|<δ$. If $x\in[x_0,x_0+h]$ then $x\in(x_0-δ,x_0+δ)$, then $|f(x)-f(x_0)|<ε$, then $|\frac{f(x)}{h}-\frac{f(x_0)}{h}|<\frac{ε}{h}$.

Therefore:
\begin{equation*}
m ( \frac{f(x)}{h},[x_0,x_0+h]) \cdot (x_0+h - x_0) ≥ \frac{f(x_0)-ε}{h} \cdot h = f(x_0)-ε
\end{equation*}
\begin{equation*}
M ( \frac{f(x)}{h},[x_0,x_0+h]) \cdot (x_0+h - x_0) ≤ \frac{f(x_0)+ε}{h} \cdot h = f(x_0)+ε
\end{equation*}

Therefore we can conclude that for a partition $P$ of $[x_o,x_0+h]$

\begin{equation*}
f(x_0)-ε< L(f,P)≤∫_{x_0}^{x_0+h}\frac{f(t)}{h}≤U(f,P)\end{equation*}
QED

Answer

What you did is correct. Here is a simpler, but not by definition, proof:

Let $F(h)=\int_{x_0}^{x_0+h}f(x)\, dx$. Since $f$ is continuous at $x_0$ by the 1st FTC, $F^{\prime}(0)=f(h)$. Therefore,,
$$\lim_{h\to 0}\frac{F(h)}{h}=\lim_{h\to 0}\frac{F(h)-F(0)}{h}=F^{\prime}(0)=f(x_0)$$
Again, this is a not by definition proof.