# Thread: Another Hard Limit Question

1. ## Another Hard Limit Question

Determine the values of a and b if $\lim_{x\to{0}}\frac{(ax+b)^{\frac{1}{3}}-2}{x} = \frac{5}{12}$

2. Originally Posted by polymerase
Determine the values of a and b if $\lim_{x\to{0}}\frac{(ax+b)^{\frac{1}{3}}-2}{x} = \frac{5}{12}$
We require that $\lim_{x\to 0}(ax+b)^{1/3} - 2 = 0$ because otherwise the limit would not exist. But that means now $b^{1/3} = 2 \implies b = 8$.
So we need to solve,
$\lim_{x\to 0}\frac{(ax+8)^{1/3} - 2}{x} = 0$.

To do this we use the fact that $f(x)=(ax+8)^{1/3}$ is differenciable at $x=0$ and so $f(x) = f(0)+xf'(0)+E(x)$ where $E(x)/x\to 0$. Thus, $f(x) = 2 + ax/12 + E(x)$.

Thus,
$\lim_{x\to 0}\frac{(ax+8)^{1/3} - 2}{x} = \lim_{x\to 0}\frac{2+ax/12 + E(x) - 2}{x} = \frac{a}{12}$.
So $a=5$