$f(x)=\{\begin{array}{cc}\frac{\sqrt[\phantom{2}]{x+2}-2}{x-2},& x<mn>2</mn>\\ \frac{1}{4},& x=2\\ \frac{15}{64}+\frac{1}{16x^2},& ,& x>2\end{array}$

$($b$)$ Use the definition of the derivative to find $f^{\text{'}}(2)$ from the left. That is$,$ compute

$f_{-}^{\text{'}}(2)=\underset{h{0}^{-}}{lim}\frac{f(2+h)-f(2)}{h}$

No marks will be given if the definition is not used.

Please help step by step and explaining why you did that!!!!

Thank you sooo much