# 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.