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Prove or disprove: If for $f:(a,b) \to \mathbb R$ the limit $\lim_{x \to a^+} \frac{f(x)-f(a)}{x-a}$ is equal to $- \infty$, then there is $a<c<b$ such that $f$ is decreasing on $(a,c)$.

It seems it should be true, since I have trouble picturing a counterexample. On the other hand, we cannot use MVT or something like that here, so I cannot prove it either.

Any ideas?

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up vote 1 down vote accepted

Let $f:[0,1)\to\Bbb{R}$ by $f(0)=0$ and

$$ f(x) = \frac{2-\cos\left( \frac{\pi}{x}\right)}{\log x}. \quad (x>0) $$

then clearly it is continuous, and

$$ \frac{f(x) - f(0)}{x-0} \leq \frac{1}{x\log x} \to -\infty $$

as $x \to 0^+$. However,

$$ f\left( \frac{1}{2n+1} \right) = -\frac{3}{\log (2n+1)} < -\frac{1}{\log (2n)} = f\left( \frac{1}{2n} \right) $$

for any large $n$. Therefore there is no such $c$.

Here is a graph of this function near the origin: W|A. You can notice how this counter-example works.

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