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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 $(a, b)$.

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See Rolle's theorem – t.b. Mar 16 '12 at 8:20
up vote 1 down vote accepted

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 $\lim\limits_{x\rightarrow a^+}f(x)=\lim\limits_{x\rightarrow b^-}f(x)=-\infty$.

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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.

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