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 $(a, b)$.
Quasi mean value theorem
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calculus
real-analysis
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1See [Rolle's theorem](http://en.wikipedia.org/wiki/Rolle%27s_theorem) – 2012-03-16
2 Answers
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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^-
A similar argument can be made when $\lim\limits_{x\rightarrow a^+}f(x)=\lim\limits_{x\rightarrow b^-}f(x)=-\infty$.
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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.