Derivative Counterexamples - Calculus I need counterexamples for the following (I guess these claims are not correct):


*

*If $ lim_{n\to \infty} n\cdot (f(\frac{1}{n}) - f(0) ) =0$ then $f$ is differentiable at $x=0$ and $f'(0)=0$ . 

*If f is defined in a neighberhood of $a$ including $a$ and differentiable at a neighberhood of $a$ (except maybe at $a$ itself), and $lim_{x\to a^- } f'(x) = lim_{x\to a^+} f'(x) $ , then $f$ is differentiable at $x=a$.

*If $f$ is diff for all $x$ and satisfies $lim_{x\to \infty } f'(x) =0 $ then there exists a number $L<\infty$ for which $ lim_{x\to \infty} f(x)= L$ 

*If $f$ is diff for all $x$ and satisfies $lim _{x\to \infty} f(x)= L $ then $lim_{x\to\infty} f'(x)=0 $ .

*If $f $ is diff at $x=0$ and $lim_{x\to 0 } \frac{f(x)}{x} =3 $ , then $f(0)=0$ and $f'(0)=3$ 
Thoughts:
5) I think this claim is correct and follows from the uniqueness of the derivative... I have no idea how to prove it, but it sounds reasonable
3) Isn't a counterexample for this is $f(x)=lnx$ ? 
4) I have tried using some trigonometric functions, but still couldn't manage to find a counterexample
2) I guess that an example for this would be a function that its derivative isn't defined at this point , but its limits do
1) have no idea... It sounds incorrect (although I guess that the other direction of the claim is correct)
Help?
Thanks !
 A: *

*There is a lot of space between $1/n$  

*The only place to go wrong is at $x=a$.  Break it there.

A: $5$. Think, if you take the limit you get $\frac{0}{0}$, what can you use then?
$4$. What would it mean if the derivative were positive or negative 'all the way' to infinity? 
$3$. Your example works just fine since $\log x$ is differentiable for all $x$ in its domain, it's slopes go to $0$, and the function grows without bound.
$2$. If a function is differentiable, then it is continuous. Think of the parabola $f(x)=x^2$ at the origin, think about the slopes as you approach $0$ from the left/right. Is there something you could do to the parabola at $0$ to make the statement untrue? 
$1$. Think derivative definition. 
A: For 1, if $\lim_{n\to\infty}n(f(1/n)-f(0))=0$ then we have $\displaystyle\lim_{n\to\infty}\frac{f(\frac{1}{n})-f(0)}{\frac{1}{n}}=0$.  It is common to assume that $n$ denotes a natural number but this was not indicated in the problem statement.  So assuming the stronger statement (i.e. that the limit is taken for $n\in\mathbb{R}$) this is equivalent to $$\lim_{h\to 0^{+}}\frac{f(h)-f(0)}{h}=0.$$  This does not mean that $\lim_{h\to 0}\frac{f(h)-f(0)}{h}$ exists.  For a counterexample think up a piecewise function.  Something like the following:
$$f(x)=\left\{\begin{array}{ll}
0& :x\geq0\\h(x)&:x<0
\end{array}\right..$$
Notice that this will satisfy the conditions of 1 regardless of what $h(x)$ is.
