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.

As the denominator tends to $0$ the limit can't be finite unless the numerator tends to $0$ are well.  So we must have $\sqrt[3]{a\cdot 0 + b} -2 = 0$ so evaluating the function at $x=0$ will yield indefinite form of $\frac 00$.
So $b=2^3 = 8$.
As the limit is indefinite form we can use L'hopital to calculate
$\lim_{x\to 0} \frac {(ax+8)^{\frac 13}}{x} = \lim_{x\to 0}  \frac {\frac 13(ax+8)^{-\frac 23}a}1= \frac {5}{12}$
So as that isn't in indefinite form. we can achieve the limit by evalutation.
So $\frac 13(a\cdot 0+8)^{-\frac 23}a=\frac 5{12}$ and
$\frac 13(a\cdot 0+8)^{-\frac 23}a= \frac a{3\sqrt[3]8^2}= \frac a{12}$
So $\frac a{12} = \frac 5{12}$ and $a=5$
