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I am trying to solve a set of problems, this one is causing my some troubles. For the first one I tried to use the $\epsilon-\delta$ definition but I couldn't solve it, I would appreciate some hints for it.

Let $f:(0,\infty) \to \mathbb{R}$ be a differentiable function such that f' is continuous and $f(x) > 0$ for all $x \in (0,\infty)$. Prove or give a counter example for each of the following statements:

  1. if $\displaystyle\lim_{x\to 0^{+}}f(x)=0$ then \displaystyle\lim_{x\to 0^{+}}f'(x) exists.

  2. if $\displaystyle\lim_{x\to\infty}f(x)=0$ then \displaystyle\lim_{x\to\infty}f'(x) = 0

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    The thing [last time](http://math.stackexchange.com/questions/34944/the-calculation-of-dimu-v-w) wasn't the use of the word "wondering". It was that you posted a formula from a source that said it was (incorrect and) derived from inclusion-exclusion, and you were "wondering" *if* the formula (though incorrect) could be derived from inclusion-exclusion. What I couldn't figure out is why you were wondering if this was the case, given that your source *said* that was the case and explained why it was the case.2011-05-28

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The answer to both is no.
(1): Consider $f(x)=x\left(2+\sin\left(\frac{1}{x}\right)\right).$ As $x\rightarrow 0$ it has to go to zero by the squeeze theorem, as $|\sin (x)|\leq 1$, and is always positive as well. However, the derivative will have the term $\frac{-1}{x}\cos\left(\frac{1}{x}\right)$ which behaves badly.

(2):

For the second one, consider $f(x)=\frac{1}{x}\left(2+\sin(x^2)\right)$ when $x\rightarrow \infty$. Certainly $f(x)\rightarrow 0$ by the squeeze theorem, and $f(x)$ is always positive since $|\sin(x)|\leq 1<2$. Lastly f^'(x)=\frac{-2}{x^2}-\frac{\sin(x^2)}{x^2}+2\cos(x), and $\cos(x)$ has no limit as $x\rightarrow \infty$.

Hope that helps,