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.
 A: 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 $a=5$
A: What about multiplying by the conjugate?
\begin{align*}
(ax+b)^{1/3} - 2 = \frac{ax + b - 8}{(ax+b)^{2/3} + 2(ax+b)^{1/3} + 4}
\end{align*}
Then, if we take $b = 8$, the proposed limit equals
\begin{align*}
\lim_{x\to 0}\frac{(ax+b)^{1/3} - 2}{x} & = \lim_{x\to 0}\frac{1}{x}\times\frac{ax + b - 8}{(ax+b)^{2/3} + 2(ax+b)^{1/3} + 4} = \frac{a}{12} = \frac{5}{12}
\end{align*}
thence we conclude that $a = 5$.
Hopefully this helps!
A: 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.
A: 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) $\implies a=5$.
A: You alos could use the binomial expansion or Taylor series
$$\sqrt[3]{a x+b}=\sqrt[3]{b}+\frac{a x}{3 b^{2/3}}-\frac{a^2 x^2}{9 b^{5/3}}+O\left(x^3\right)$$
$$\frac{\sqrt[3]{a x+b} - 2}{x}=\frac{\sqrt[3]{b}-2}{x}+\frac{a}{3 b^{2/3}}-\frac{a^2 x}{9
   b^{5/3}}+O\left(x^2\right)$$ So, $$\sqrt[3]{b}-2=0 \implies b=8$$
$$\frac{a}{3 b^{2/3}}=\frac{a}{12}=\frac{5}{12}\implies a=5$$
