Limit $\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x}$ 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
$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12}$
This is what I tried with better results.
For $b$:
$$\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12} $$
$$\sqrt[3]{a x+b} - 2 = x\frac{5}{12} $$
$$\underset{x\to 0}{\text{lim}} \sqrt[3]{a x+b} - 2 = \underset{x\to 0}{\text{lim}} x\frac{5}{12} $$
$$\sqrt[3]{b} - 2 = 0 $$
$$\sqrt[3]{b}  = 2 $$
$$ b = 8$$
For $a$:
$$a x+8 = (x\frac{5}{12} + 2)^3$$
$$a x+8 = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x+8$$
$$a x = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x$$
$$a = \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$\underset{x\to 0}{\text{lim}} a =\underset{x\to 0}{\text{lim}}  \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$a = 5 $$
But the limit $$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{5x+8} - 2}{x}$$ doens't go to $\frac {5}{12}$.
May someone help.

Actually, the limit
$\lim\limits_{x\to 0}\frac{(5x+8)^{1/3}-2}{x}$ (of the form $\frac{0}{0}$)
$=\lim\limits_{x\to 0}\frac{5.\frac{1}{3}(5x+8)^{-2/3}}{1}$, by L' Hospital's rule (since the limiting value of both numerator and denominator is 0 and the first derivative of the denominator w.r.t. $x$ is non-zero, the rule is applicable).
$=\frac{5}{12}$
Also, letting $y^3=ax+b$, we have the limit $L=\lim\limits_{y\to b^{1/3}}a.\frac{y-2}{y^3-b}$, the denominator tends to $0$, the limit exists iff numerator tends to $0$ too, $\implies b=8$.
Now, $L=\lim\limits_{y\to 2} \frac{a}{y^2+2y+4}=\frac{a}{12}=\frac{5}{12}$ (given) $\implies a=5$.
