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


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*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 !

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.
