Another form of the Sandwich theorem (for derivatives in dimension $1$) Here is the theorem :
"Let $I\subseteq \mathbb{R}$ an interval which contains $a\in \mathbb{R}$. Let $M$ and $m$ two functions defined on $I$, differentiable at $a$ and $f$ a function defined on $I$ which takes value in $\mathbb{R}$.
If : 
$i)\ \forall \ x \in I , m(x)\le f(x)\le M(x)$
$ii)\ m(a) = f(a) = M(a)$
$iii)\ m'(a)= M'(a)$
Then $f$ is differentiable at $a$ and we have $f'(a)=m'(a)=M'(a)$"
Here is my proof :
I consider the function : $x\mapsto \frac{f(x)-f(a)}{x-a}$ with $x\ne a$
Then $\forall \ x \in I \setminus \{a\}$ by using $i)$ we have directly :
$\frac{m(x)-f(a)}{x-a} \le \frac{f(x)-f(a)}{x-a} \le \frac{M(x)-f(a)}{x-a}$ if $x-a>0$ (for the other case we just have to change the sign of the inequality)
But by $ii)$ we have : $m(a)=f(a)=M(a)$ so we obtain :
$\frac{m(x)-m(a)}{x-a} \le \frac{f(x)-f(a)}{x-a} \le \frac{M(x)-M(a)}{x-a}$
$\Rightarrow$
$m'(a) \le \lim \limits_{x\to a} \frac{f(x)-f(a)}{x-a} \le M'(a)$
Then by using $iii)$ we have : $m'(a) \le \lim \limits_{x\to a} \frac{f(x)-f(a)}{x-a} \le m'(a)$ or $M'(a) \le \lim \limits_{x\to a} \frac{f(x)-f(a)}{x-a} \le M'(a)$
So we prove that $f$ is differentiable at $a$ (by the Squeeze theorem the limit gives $m'(a)$ or $M'(a)$) and we have $f'(a)=m'(a)=M'(a)$
The theorem is proved.
Am I right ? Thanks in advance.
 A: NOTE:  I originally thought the below was a counter-example, but it's not actually one.  Reasons below.

The theorem as stated is false.  Counter-example, defined on the interval $I = [-1, 1]$:
$$
m(x) = -x^2 \qquad M(x) = x^2 \qquad f(x) = \begin{cases} x^2 \sin \left( \frac{1}{x^2} \right) & x \neq 0 \\ 0 & x = 0 \end{cases}
$$
Condition (i) is satisfied everywhere on the interval, and conditions (ii) and (iii) are satisfied at $a = 0$.  But $f(x)$ is not differentiable at $a = 0$.

My error above (as pointed out by Jhilbert in the comments) is that $f'(0) = \lim_{x \to 0} f(x)/x$ does in fact exist, and is equal to zero.  However, $\lim_{x \to 0} f'(x)$ does not exist.  If this "counterexample" shows anything, it is that the above conditions are insufficient to guarantee that $f'(x)$ has a well-defined limit at $a$, even if $m(x)$ and $M(x)$ are smooth functions there.
A: It's correct. just a formal thing: you can't move to the inequality 
$$
m^\prime(a)\leq\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}\leq M^\prime(a)
$$
Because you haven't proved that the limit exists. you need to move from the first inequality straight to the conclusion.
