Are limits commutative? Generally speaking, is the following true:
$$\lim_{x\to a}f'(x)=\lim_{x\to a}\left(\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\right)=\lim_{h\to 0}\left(\lim_{x\to a}\frac{f(x+h)-f(x)}{h}\right)$$
Your statement boils down to the continuity of $f'.$ 

However, the derivative, even if it exists for any point $x\in\mathbb{R},$ does not have to be continuous. A classic counterexample is $f(x)=x^2\sin(1/x)$ for $x\neq 0,$ and $f(0)=0.$
If the limit of derivative exists at a point $a$, then function can be approximated by a line at points really close to left of $a$, now if the limit of derivative exists, then another line with same slope as previous line approximates the points really close to right of $a$. That was about LHS of the equation, now coming to RHS, your first limit means you approach $a$ from any side, and fix an $x$ really close to $a$, with second limit you are talking about derivative at that $x$, which may not exist, but as you get closer to $a$,    the derivative is defined at those points close to $a$, because your LHS(limit of derivative exists), meaning derivative exists around a neighborhood of $a$. 

As for examples for equality,you can have $y=x$, at non zero $x$, and make this function discontinuous at zero.