# questions about limits and derivatives

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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You have been in this site for about 8 months according to your profile. Are you not yet aware that it is impolite to post in the imperative mode, as if you were assigning homework? Or that the best steps to take when posting a homework problem is to add what it is you are having trouble with or what you have tried so far? – Arturo Magidin May 28 '11 at 0:49
Arturo is right. There is no excuse if you have been here for a long time. So, what have you tried so far? (Etc...) – Eric Naslund May 28 '11 at 0:56
Sorry for being rude, I'm editing the post right now. And I'm avoiding to post something else like "wondering" because last time Arturo used it against me :'( – Vicfred May 28 '11 at 0:59
The thing last time 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. – Arturo Magidin May 28 '11 at 1:42

(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.
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 $x\rightarrow \infty$.