Answer
$a = 5$
$b = 8$
Work Step by Step
$\lim\limits_{x \to 0}\frac{\sqrt[3] {ax+b}-2}{x} = \frac{5}{12}$
Note that $\lim\limits_{x \to 0}\frac{\sqrt[3] {ax+b}-2}{x} = f'(0)$ when $f(x) = \sqrt[3] {ax+b}$ and $f(0) = 2$
Since $f(0) = 2$:
$f(x) = \sqrt[3] {ax+b}$
$f(0) = \sqrt[3] {a(0)+b} = 2$
$\sqrt[3] {b} = 2$
$b = 8$
We can find the value of $a$:
$f(x) = \sqrt[3] {ax+b}$
$f'(x) = \frac{a}{3~(ax+b)^{2/3}}$
$f'(0) = \frac{a}{3~(a(0)+8)^{2/3}} = \frac{5}{12}$
$\frac{a}{3~(8)^{2/3}} = \frac{5}{12}$
$\frac{a}{12} = \frac{5}{12}$
$a = 5$
