Showing $g'(x_0)$ exists and equals $f'(x_0)$ and $h'(x_0)$ if $f(x_0)$=$h(x_0)$ I have three functions, $f,g,h$ that all map from $[a,b]$ to $\mathbb{R}$. I know the following:
$$f(x) \leq g(x) \leq h(x)\ \forall x\in [a,b]  $$
$$x_0 \in (a,b)$$
$$f(x_0)=h(x_0)$$
$$h'(x_0) \ \& \ f'(x_0) \ exist$$
I want to show that $g'(x_0)$ exists and that $g'(x_0)=f'(x_0)=h'(x_0)$
Any suggestions? I'm really lost. I tried applying the definition of something being differentiable at a point but made no progress.
 A: Consider the definition:
$g$ is differentiable at $x_0$ if 
$$
\lim_{x\to x_0}\frac{g(x_0)-g(x)}{x-x_0}
$$
exists or equivalent if
$$
\liminf_{x\to x_0}\frac{g(x_0)-g(x)}{x-x_0}=\limsup_{x\to x_0}\frac{g(x_0)-g(x)}{x-x_0}.
$$
or equivalent if
$$
\lim_{\substack{x\to x_0\\x>x_0}} \frac{g(x_0)-g(x)}{x_0-x} \text{ and }
\lim_{\substack{x\to x_0\\x<x_0}} \frac{g(x_0)-g(x)}{x_0-x}
$$
exist and are equal.
Remark:
The limit
$$
\lim_{\substack{x\to x_0\\x>x_0}} \frac{g(x_0)-g(x)}{x_0-x}
$$
exists if
$$
\liminf_{\substack{x\to x_0\\x>x_0}}\frac{g(x_0)-g(x)}{x-x_0}=\limsup_{\substack{x\to x_0\\x>x_0}}\frac{g(x_0)-g(x)}{x-x_0}.
$$
and analogue for $<$ insteat of $>$.
Proof:
Now we have $g(x_0)=f(x_0)=h(x_0)$. If $x<x_0$ we get
$$
\frac{h(x_0)-h(x)}{x_0-x}=
\frac{g(x_0)-h(x)}{x_0-x}\leq 
\frac{g(x_0)-g(x)}{x_0-x}=
\frac{f(x_0)-g(x)}{x_0-x}\leq
\frac{f(x_0)-f(x)}{x_0-x}
$$
and for $x>x_0$ we get
$$
\frac{h(x_0)-h(x)}{x_0-x}=
\frac{g(x_0)-h(x)}{x_0-x}\geq 
\frac{g(x_0)-g(x)}{x_0-x}=
\frac{f(x_0)-g(x)}{x_0-x}\geq
\frac{f(x_0)-f(x)}{x_0-x}
$$
Since $f$ and $h$ are differentiable at $x_0$ we get
\begin{align*}
f'(x_0)&=\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x_0-x}=
\lim_{\substack{x\to x_0\\x>x_0}}\frac{f(x_0)-f(x)}{x_0-x}
\\&\leq 
\lim_{\substack{x\to x_0\\x>x_0}}\frac{h(x_0)-h(x)}{x_0-x}
=\lim_{x\to x_0}\frac{h(x_0)-h(x)}{x_0-x}=h'(x_0)
\end{align*}
and
\begin{align*}
f'(x_0)&=\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x_0-x}=
\lim_{\substack{x\to x_0\\x<x_0}}\frac{f(x_0)-f(x)}{x_0-x}
\\&\geq 
\lim_{\substack{x\to x_0\\x<x_0}}\frac{h(x_0)-h(x)}{x_0-x}
=\lim_{x\to x_0}\frac{h(x_0)-h(x)}{x_0-x}=h'(x_0)
\end{align*}
From $f'(x_0)\leq h'(x_0)$ and $f'(x_0)\geq h'(x_0)$ we get $f'(x_0)=h'(x_0)$ and you can conclude
$$
\lim_{x\to x_0}
\frac{h(x_0)-h(x)}{x_0-x}=\lim_{x\to x_0}
\frac{g(x_0)-g(x)}{x_0-x}=\lim_{x\to x_0}
\frac{f(x_0)-f(x)}{x_0-x}
$$
If you think, the last argument is to sloppy you can extend the last argument following way:
Since $
\frac{h(x_0)-h(x)}{x_0-x}\leq \frac{g(x_0)-g(x)}{x_0-x}$ for all $x\in(a,x_0)$ you get
$$
\liminf_{\substack{x\to x_0\\x<x_0}}\frac{h(x_0)-h(x)}{x_0-x}\leq 
\liminf_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x} 
$$
and $\frac{g(x_0)-g(x)}{x_0-x}\leq \frac{f(x_0)-f(x)}{x_0-x}$ for all $x\in(a,x_0)$ imply
$$
\limsup_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x}\leq 
\limsup_{\substack{x\to x_0\\x<x_0}}\frac{f(x_0)-f(x)}{x_0-x} 
$$
These are valid since $\liminf$ and $\limsup$ can be defined even if the limit doesn't exists or the term is unbounded.
And now you can use that $f$ and $g$ are differentiable at $x_0$ which imply
$$
h'(x_0)=\lim_{x\to x_0}\frac{h(x_0)-h(x)}{x_0-x}=
\liminf_{\substack{x\to x_0\\x<x_0}}\frac{h(x_0)-h(x)}{x_0-x}
$$
and
$$
f'(x_0)=\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x_0-x}=
\limsup_{\substack{x\to x_0\\x<x_0}}\frac{f(x_0)-f(x)}{x_0-x}
$$
So we get
$$
h'(x_0)\leq 
\liminf_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x} \leq
\limsup_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x}\leq 
f'(x_0)
$$
Finally we use $f'(x_0)=h'(x_0)$ and conclude
$$
h'(x_0)=\liminf_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x}=
\limsup_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x}=f'(x_0)
$$
Therefore $\lim_{\substack{x\to x_0\\x<x_0}} \frac{g(x_0)-g(x)}{x_0-x}$ exists and equals $f'(x_0)$ and $h'(x_0)$. Analogue we can prove that $\lim_{\substack{x\to x_0\\x>x_0}} \frac{g(x_0)-g(x)}{x_0-x}$ exists and equals $f'(x_0)$ and $h'(x_0)$. Together we get that $g$ is differentiable at $x_0$ and $g'(x_0)$ equals $f'(x_0)$ and $h'(x_0)$.
