Let $f : \mathbb{R} \to \mathbb{ R}$ be such that $f' (x)$ exists Let $f : \mathbb{R} \to \mathbb{ R}$ be such that $f' (x)$ exists for all non zero $x$ and $\lim_{x\to 0} f' (x) = 0$.
Then
(i) $f$ is continuous but not differentiable at $0$.
(ii) $f$ is differentiable at $0$ and $f' (0) = 0.$
(iii) $f$ has either a local maximum or a local minimum at $0$.
(iv) None of the above.
I am totally clueless. Thank you for help and discussion.
 A: Take $f(x) = \begin{cases} 1 & x >0 \\ 0 & x = 0 \\ -1 & x<0 \end{cases}$.
Then $f$ is not continuous at $x=0$ (and hence not differentiable).
$f$ has neither a local maximum or minimum at $x=0$.
Hence the answer is (iv) None of the above.
A: IF $f$ is continuous at $0$, it is easy to prove it is also differentiable.
Let $x \neq 0$. Then, by the MVT, there exists some $c_x$ between $0$ and $x$ so that
$$\frac{f(x)-f(0)}{x-0}=f'(c_x) $$
Now
$$\lim_{x \to 0}f'(x)=0 \Rightarrow \lim_{x \to 0}f'(c_x)=0 \Rightarrow \frac{f(x)-f(0)}{x-0}=0 \,.$$
This shows that $(i)$ is False. 
(iii) $f(x)=x^3$.
A: In my answer I suppose that $f$ is continuous at $0$
Let $x>0$ then by the mean value theorem there's $\xi_x\in(0,x)$ s.t.
$$f(x)-f(0)=xf'(\xi_x)$$
so by passing to the limit $x\to0$ so $\xi_x\to 0$ and since $\lim_{x\to 0}f'(x)$ exists  we have
$$\lim_{x\to0}f'(\xi_x)=0=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=f'(0)$$
hence $f'(0)=0$.
Added In the case where $f$ isn't continuous at $0$ you can see the answer of cooper.hat for a counterexample.
