calculus – Limit $underset{xto 0}{text{lim}}frac{sqrt[3]{a x+b} – 2}{x}$ equals to $frac{5}{12}$

A friend of mine asked this question to me. It seems it’s from Stewart.

Find the values of a and b such that
$underset{xto 0}{text{lim}}frac{sqrt(3){a x+b} – 2}{x} = frac{5}{12}$

This is what I tried with better results.

For b:

$frac{sqrt(3){a x+b} – 2}{x} = frac{5}{12} $

$sqrt(3){a x+b} – 2 = xfrac{5}{12} $

$underset{xto 0}{text{lim}} sqrt(3){a x+b} – 2 = underset{xto 0}{text{lim}} xfrac{5}{12} $

$sqrt(3){b} – 2 = 0 $

$sqrt(3){b} = 2 $

$ b = 8$

For a:

$a x+8 = (xfrac{5}{12} + 2)^3$

$a x+8 = frac{125 x^3}{1728}+frac{25 x^2}{24}+5 x+8$

$a x = frac{125 x^3}{1728}+frac{25 x^2}{24}+5 x$

$a = frac{125 x^2}{1728}+frac{25 x}{24}+5$

$underset{xto 0}{text{lim}} a =underset{xto 0}{text{lim}} frac{125 x^2}{1728}+frac{25 x}{24}+5$

$ a = 5 $

But the limit $underset{xto 0}{text{lim}}frac{sqrt(3){5x+8} – 2}{x}$ doens’t goes to $frac {5}{12}$.

May someone help.