Solve for constants: Derivatives using first principles 
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*Question
Find the values of the constants $a$ and $b$ such that $$\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \frac{5}{12}$$



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*My approach


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*Using the definition of the derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

*I view limit as a derivative of a function $f$ at some value, let's call that value $c$, as follows :$$f'(c) = \lim_{x\to 0}\frac{f(c + x) - f(c)}{x} = \frac{\sqrt[3]{ax + b} - 2}{x} = \frac{5}{12}$$

*Now I deduce the following: $$f(c + x)  = \sqrt[3]{ax + b}$$ and $$f(c) = 2$$

*Use limits as follows: $$\lim_{x\to 0} f(c + x) = f(c) = 2$$ that is, $$\lim_{x\to 0} f(c + x) = \lim_{x\to 0} \sqrt[3]{ax + b} = \sqrt[3]{b} = 2$$ now solve for $b$, $$\sqrt[3]{b} = 2 \Leftrightarrow b = 8$$

*Since I know that $$f(c + x) = \sqrt[3]{ax + 8}$$, I can solve for $a$, which is $$a = \frac{[f(c+x)]^3 - 8}{x} = \frac{[f(c+x)]^3 - [f(c)]^3}{x}$$

*Let $g(x) = [f(x)]^3$, such that $$g'(x) = 3\cdot [f(x)]^2 \cdot f'(x)$$
so $$g'(c) = 3\cdot [f(c)]^2 \cdot f'(c) = 3 \cdot 4 \cdot \frac{5}{12} = 5$$

*Rephrase $g'(c)$ using first principles such that $$g'(c) = \lim_{x \to 0}\frac{g(c + x)- g(c)}{x}= \lim_{x \to 0}\frac{[f(c + x)]^3 - [f(c)]^3}{x} = \lim_{x \to 0} a = 5$$

*Since $a$ is a constant, $\lim_{x \to 0} a = a$, that is, $$a = 5$$

*My solution: $b = 8, a = 5$.

Please have a look at my approach and give me any hints/suggestions regarding the solution and/or steps taken.
 A: It is fine. However, there is a short proof if you know this 

$(1+x)^\frac1n=1+\frac1nx+o(x)$, where $\lim_{x\to 0}\frac{o(x)}{x}=0.$

So $$\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \lim_{x \to 0}\frac{b^\frac13 \sqrt[3]{\frac a bx + 1}-2}{x}=\lim_{x \to 0}\frac{(b^\frac13-2)+  \frac13 b^\frac13 \frac abx+b^\frac13o(x)}{x}=\frac{5}{12}.$$
So, $b^\frac13-2=0$ and $\frac13 b^\frac13 \frac ab=\frac{5}{12}$.
A: Your approach looks fine, and it is explained well.  Here is a variant of your solution, that uses similar ideas:
1) $\displaystyle\lim_{x\to0}(\sqrt[3]{ax+b}-2)=\lim_{x\to0}\bigg(\frac{\sqrt[3]{ax+b}-2}{x}\cdot x\bigg)=\frac{5}{12}\cdot 0=0,\;\;\;$ so $\sqrt[3]{b}=2$ and $b=8$.
2) Let $g(x)=\sqrt[3]{ax+8}.\;\;\;$  Then $\displaystyle g^{\prime}(0)=\lim_{x\to0}\frac{g(x)-g(0)}{x-0}=\frac{\sqrt[3]{ax+8}-2}{x}=\frac{5}{12}$,
so $g^{\prime}(x)=\frac{1}{3}(ax+8)^{-2/3}\cdot a\implies\frac{1}{3}\cdot\frac{1}{4}a=\frac{5}{12}\implies a=5$.
A: Using
$$ a^3-b^3=(a-b)(a^2+ab+b^2, $$
one has
\begin{eqnarray}
\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \lim_{x \to 0}\frac{ax+b-8}{x[(\sqrt[3]{ax + b})^2+2\sqrt[3]{ax + b}+4]}.
\end{eqnarray}
Thus if the limit exists, one must have $b=8$, under which it is easy to get
\begin{eqnarray}
\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} &=& \lim_{x \to 0}\frac{ax+b-8}{x[(\sqrt[3]{ax + b})^2+2\sqrt[3]{ax + b}+4]}\\
&=&\frac{a}{[(\sqrt[3]{8})^2+2\sqrt[3]{8}+4]}=\frac{a}{12}.
\end{eqnarray}
Thus $a=5$.
