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.

It appears from your working that you had the right idea (at least at the start) in mind and the issue was more related to the use of limit laws.
Let us fix your approach as follows. We have $$\lim_{x\to 0}(\sqrt[3]{ax+b}-2)=\lim_{x\to 0}x\cdot\frac {\sqrt [3]{ax+b} - 2}{x}=0\cdot\frac{5}{12}=0$$ which means that $$\sqrt[3]{b}-2=0$$ or $b=8$.
To find $a$ we need a bit more work. Observe that if $a=0$ then the limit in question becomes $0$. Hence $a\neq 0$ and let us put $t=ax+b$ so that $t\to 8$ as $x\to 0$. The limit in question can be written as $$\lim_{t\to 8}\frac{t^{1/3}-8^{1/3}}{t-8}\cdot\frac{t-8}{x}=\frac{5}{12}$$ ie $$\frac{1}{3}\cdot 8^{-2/3}\cdot a =\frac{5}{12}$$ or $a=5$.
To summarize $a=5,b=8$. We have use the standard limit $$\lim_{t\to c} \frac{t^n-c^n} {t-c} =nc^{n-1}$$ (with $c=8,n=1/3$) above.
You should also note that if limit of a function $f(x) $ is $L$ then it does not necessarily hold that $f(x) =L$. This is one crucial mistake which you made while finding $a$.

Food for thought: Why did I specifically eliminate the option $a=0$? The argument after that probably works without this assumption.
