Quasi mean value theorem 
Let $f$ be differentiable on $(a,b)$ and $\lim\limits_{x\to{a^+}}f(x)=\lim\limits_{x\to{b^-}}f(x)$. Prove $f\,'$ has at least one zero on $(a, b)$.

 A: If the one-sided limits are finite, then define $f$ at $a$ and $b$ to be the common  value of the limit (this extended function will be continuous on $[a,b]$). Then use Rolle's Theorem. 
If the one sided limits are infinite, there's a bit more work to do. Here, you could mimic the proof of Rolle's Theorem. 
For example, in the case where
$\lim\limits_{x\rightarrow a^+}f(x)=\lim\limits_{x\rightarrow b^-}f(x)=\infty$:
Let $d\in(a,b)$. Choose $a^+>a$ close to $a$ and $b^-<b$ close to $b$ such that
$d\in(a^+,b^-)$ and both $f(a^+)$ and $f(b^-)$ exceed $f(d)$. $f$ is continuous on $[a^+,b^-]$, and so has a minimum value over $[a^+,b^-]$ at some point $c\in(a^+,b^-)$ (the minimum cannot occur at the endpoints). Since $f$ is differentiable over $(a^+,b^-)$, we have $f'(c)=0$.
A similar argument can be made when $\lim\limits_{x\rightarrow a^+}f(x)=\lim\limits_{x\rightarrow b^-}f(x)=-\infty$.
A: Set $f(a)=f(b)=m$, where m is the limit. Then use mean value theorem on the close interval $[a,b]$. I think it might be some homework.
