First, note that direct substitution for all of the choices result in the indeterminate form. Let's go through the choices one by one.
$$ \lim\limits_{x\to 0} \frac{f(x)-f(0)}{x} =0 $$
This is simply the limit definition of the derivative at \(x=0\). Recall:
$$ f'(a)=\lim\limits_{x\to a} \frac{f(x)-f(a)}{x-a} $$
The derivative at \(x=0\) does not exist since according to the graph, the left and right hand limits are \(-1\) and \(1\). Therefore, the first option is false.
$$\lim\limits_{x\to 0} \frac{f(x)-f(-x)}{2x} =0 $$
Since the graph is an even function, \(f(x)=f(-x)\). The limit can be simplified:
$$\lim\limits_{x\to 0} \frac{f(x)-f(-x)}{2x} = \lim\limits_{x\to 0} \frac{f(x)-f(x)}{2x} $$
$$ = \lim\limits_{x\to 0} \frac{0}{2x} = \lim\limits_{x\to 0} 0 = 0 \checkmark$$
$$\lim\limits_{x\to 4} \frac{f(x)-f(4)}{x-4} = -1 $$
This is the definition of the derivative at \(x=4\):
$$ f'(a)=\lim\limits_{x\to a} \frac{f(x)-f(a)}{x-a} $$
$$ f'(4)=\lim\limits_{x\to a} \frac{f(x)-f(4)}{x-4} $$
The slope is \(-1\) at this point. \(\checkmark\)
$$\lim\limits_{x\to 2} \frac{f(x)-f(2)}{x-2} \text{does not exist.} $$
This is the definition of the derivative at \(x=2\)
$$ f'(a)=\lim\limits_{x\to a} \frac{f(x)-f(a)}{x-a} $$
$$ f'(2)=\lim\limits_{x\to 2} \frac{f(x)-f(2)}{x-2} $$
The derivative does not exist at \(x=2\) since the left and right hand limits are \(1\) and \(-1\). \(\checkmark\)