What is the value of $ab$ if $\lim_{x\to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \frac{7}{12}$?

Question

Here's the question:

Let $a$ and $b$ be real numbers such that $$\lim_{x\to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \frac{7}{12}$$ What's the value of $a\cdot b$?

This question was on a military exam here on Brazil, and a YouTube solution for that went like this:

For that limit to exist, the indetermination $\frac{0}{0}$ is demanded, so we can use the L'Hopital rule. Then, we gotta have $ax + b = 8$, and as $x\to 0$, we have $b = 8$. Therefore, via L'Hopital:

$$\begin{align}\displaystyle\lim_{x\to 0}\frac{\sqrt[3]{ax + 8}-2}{x} &= \displaystyle\lim_{x\to 0}\frac{(\sqrt[3]{ax + 8}-2)'}{(x)'}\\ &= \displaystyle\lim_{x\to 0}\frac{\frac{1}{3}\cdot (ax+ 8)^{-\frac{2}{3}}\cdot a}{1}\\ &=\frac{1}{3}\cdot 2^{-2}\cdot a\\ &=\frac{a}{12}. \end{align}$$

Thus, $a = 7$ and then $a\cdot b = 7\cdot 8 = 56.$

Now, here's my doubt. In order for the limit $\displaystyle\lim_{x\to 0}\frac{\sqrt[3]{ax + 8}-2}{x}$ to exist and equals $\frac{7}{12}$, is it really necessary to force the $\frac{0}{0}$ indetermination? I mean, is it a necessary and a sufficient condition?

Answer 1

Without L'Hospital, considering $$y=\frac{\sqrt[3]{ax + b}-2}{x} $$ using Taylor expansion around $x=0$ or binomial theorem $$\sqrt[3]{ax + b}=\sqrt[3]{b}+\frac{a x}{3 b^{2/3}}-\frac{a^2 x^2}{9 b^{5/3}}+O\left(x^3\right)$$ making $$y=\frac{\sqrt[3]{b}-2+\frac{a x}{3 b^{2/3}}-\frac{a^2 x^2}{9 b^{5/3}}+O\left(x^3\right)}x=\frac{\sqrt[3]{b}-2}{x}+\frac{a}{3 b^{2/3}}-\frac{a^2 x}{9 b^{5/3}}+O\left(x^2\right)$$

So, $\sqrt[3]{b}-2 \implies b=8$ and $\frac{a}{3 b^{2/3}}=\frac 7 {12}\implies a=7$.

Answer 2

Without De l'Hopital, you could simply go this way (recalling the factorization $a^3-b^3 = (a-b)(a^2+ab+b^2))$.

$$\lim_{x\to 0}\frac{\sqrt[3]{ax+b}-2}{x}=\lim_{x\to 0}\frac{ax+b-8}{x\left(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\right)}.$$

Again here, as in comments, we need the numerator to be $0$ in $x=0$, which means $b=8$. Therefore the limit becomes

$$\lim_{x\to 0}\frac{a}{\sqrt[3]{(ax+8)^2}+2\sqrt[3]{ax+8}+4}=\frac{a}{12}.$$

The result follows.