Say I have three functions $f_-(x)$, $f_+(x)$, and $f(x)$ that obey: \begin{align} f_-(x) &\le f(x) \le f_+(x), \\ f_+'(x) &\le f_-'(x), \\ f_0 & = \lim_{x \rightarrow x_0} f_-(x) = \lim_{x \rightarrow x_0} f_+(x),\ \mathrm{and} \\ f'_0 & = \lim_{x \rightarrow x_0} f_-'(x) = \lim_{x \rightarrow x_0} f_+'(x), \end{align} for all $x$ in some (possibly infinite) open interval $[a,b]$ ($x_0$ in the same interval). Both the upper and lower bounds are at least $C_1$ continuous. Naturally, this is sufficient to show that $\lim_{x\rightarrow x_0} f(x) = f_0$ by the sandwich (squeeze) theorem.
I want to also show that $f'(x)$ is similarly sandwiched. Are the above properties sufficient to show that $\lim_{x\rightarrow x_0} f'(x) = f'_0$? If not, is it sufficient to require that $f(x)$ also be $C_1$ continuous to get the desired result? If not, what additional limitations/properties need to be imposed on $f(x)$ to ensure the desired result?