Hint required - Find values of $a$ and $b$ such that the limit equals $\frac{5}{12}$. Find the values of $a$ and $b$ such that $$\lim_{x\to 0} \frac{\sqrt[3]{ax+b}-2}{x}= \frac{5}{12}.$$
Can someone please provide me with a hint on how to approach this question?
 A: Can you link your expression with the definition of the derivative of a differentiable function $f$ at a point $c$?
$\lim_{h \rightarrow 0} \frac{f(c+h) - f(c)}{h}$
A: $$\lim_{x\to 0} \frac{\sqrt[3]{ax+b}-2}{x}= \frac{5}{12}$$
An approach without using L'Hôpital's rule:
Since the denominator approaches zero as $x$ approaches zero, the numerator must also approach zero as $x$ approaches zero for the limit to be finite, i.e.,
$$\tag{1} \lim_{x\to0}\sqrt[3]{ax+b}-2=0
\implies b^\frac13 - 2 = 0$$
Now, consider the numerator:
$$\begin{align}\sqrt[3]{ax+b}-2 &= \sqrt[3]{\left(\frac{ax}{b}+1\right)b}-2 \\
&= b^{\frac13} \left(1+\frac{ax}{b}\right)^{\frac13} - 2 \\
\end{align}$$
Using the binomial approximation, i.e., $(1+x)^n \approx1+nx$ for small $x$, we get
$$ b^{\frac13}\left(1+\frac{ax}{3b}\right) - 2 = \underbrace{(b^\frac13-2)}_{0} + \frac{ax}{3b^\frac23} = \frac{ax}{3b^\frac23}$$
Thus,
$$ \tag{2} \lim_{x\to 0} \frac{\sqrt[3]{ax+b}-2}{x} = \lim_{x\to 0} \frac{\frac{ax}{3b^{2/3}}}{x} = \frac{a}{3b^\frac23} = \frac{5}{12}$$
It is now easy to calculate $a$ and $b$ from $(1)$ and $(2)$
