Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.