Limit without l'hopital

Question

Find a and b such $$\lim_{x\to 0}\frac{(ax+b)^\frac{1}{3} - 2}{x} = \frac{5}{12}$$

My Solution : I know that $b$ has to be $8$. In order to have $\displaystyle \frac{0}{0}$. But for finding a, I have to factorize and cross out $x$, which I didn't get how. I've done this by applying L'hopital rule and I get $a$ is 5. But I don't know how to do it without L'hopital rule. Can someone help me? Thanks.

Answer 1

Let $ax+8 = A^3$ and $2=B\;,$ Then $$\displaystyle ax = A^3-B^3\Rightarrow x=\frac{A^3-B^3}{a}$$

So we get $$\displaystyle \lim_{x\rightarrow 0}\frac{(A-B)\cdot a}{(A^3-B^3)} = \lim_{x\rightarrow 0}\frac{a}{A^2+B^2+AB} = \lim_{x\rightarrow 0}\frac{a}{(ax+8)^{\frac{2}{3}}+4+2\cdot (ax+8)^{\frac{1}{3}}}$$

So we get $$\displaystyle \frac{a}{12} = \frac{5}{12}$$ (Given)

So we get $a = 5$

Answer 2

Use the taylor series

$$(ax+8)^{\frac{1}{3}}=2+\frac{ax}{12}+O(x^2)$$

to get $a=5$.