Definition of a Derivative problem 
$$f(x) =\begin{cases}
2x-2 & x< 3\\
2x-4 & x\geq 3
\end{cases}$$


6: Let $f$ be the piecewise linear function defined above. Which of the following statements are true?


I: $$\lim_{h\to 0^-}\frac{f(3+h)-f(3)}{h}=2$$


II: $$\lim_{h\to 0^+}\frac{f(3+h)-f(3)}{h}=2$$


III: $f'(3)=2$


(A): None   (B): II only   (C): I and II only    (D): I, II, and III

The answer key for this problem says that the answer is B, but I can't seem to understand why. Shouldn't it be C? The derivative of this function when $x$ is just less than and just greater than $3$ is $2$ either way, right?
 A: It is true that for any $\epsilon > 0$ you have $f'(3 - \epsilon) = f'(3 + \epsilon) = 2$. However, it is very important to look at the limits actually described, noting in particular that by the function's definition $f(3) = 2$. So if we calculate the limit in I:
$$\begin{eqnarray}\lim_{h \rightarrow 0^-} \frac{f(3 + h) - f(3)}{h} & = & \lim_{h \rightarrow 0^-} \frac{(2(3 + h) - 2) - 2}{h} && \text{Since } h < 0, 3+h<3 \\
& = & \lim_{h \rightarrow 0^-} \frac{6 + 2h - 4}{h} \\
& = & \lim_{h \rightarrow 0^-} \frac{2h + 2}{h}
\end{eqnarray}$$
and this limit does not exist. Which means that $f$ does not have a left derivative at $x = 3$, and so it has no derivative.
A: $f(3)=2x-4$; so when you plug this into the limits in both I and II, II gives you $2$ as expected but you get $$\lim_{h\rightarrow0^{-}}\dfrac{6+2h-2-6+4}{h}=\lim_{h\rightarrow0^{-}}\dfrac{2h+2}{h}$$ for I, which doesn't quite reduce to $2$.
I'm guessing what you did was plug in $f(3)=2\times 3-2$ for I and find the limits to be the same.
