Answer
$\lim\limits_{h \to 0}\frac{f(x+h)-f(x-h)}{2h} = f'(x)$
The diagram shows that the slope of the line connecting the points $(x-h, f(x-h))$ and $(x+h, f(x+h))$ approaches the slope of the graph at $x$ as $h\to 0$
Work Step by Step
$\lim\limits_{h \to 0}\frac{f(x+h)-f(x-h)}{2h} = \frac{0}{0}$
We can apply L'Hospital's Rule.
$\lim\limits_{h \to 0}\frac{f'(x+h)-(-1)f'(x-h)}{2} = \frac{2f'(x)}{2} = f'(x)$
The diagram shows that the slope of the line connecting the points $(x-h, f(x-h))$ and $(x+h, f(x+h))$ approaches the slope of the graph at $x$ as $h\to 0$
