$$f(x) = \begin{cases} \frac{\sqrt{x+2}-2}{x-2}, & x < 2, \\ \frac{1}{4}, & x = 2, \\ \frac{15}{64} + \frac{1}{16x^2}, & x > 2. \end{cases}$$
(b) Use the definition of the derivative to find $f'(2)$ from the left. That is, compute
$$f'_{-}(2) = \lim_{h\to 0^-} \frac{f(2+h) - f(2)}{h}.$$
No marks will be given if the definition is not used.