Suppost that f is a differentiable function such that $\displaystyle f'(8)=2$. Without using L'Hospital's Rule, find the value of $\displaystyle \displaystyle\lim_{x\to 8}\frac{f(x)-f(8)}{x^{\frac{1}{3}}-2}$
do you not see the limit we want in there?
$\displaystyle 2 = \lim_{x \to 8} \frac {f(x) - f(8)}{\left( x^{\frac 13} - 2 \right) \left( x^{\frac 23} + 2x^{\frac 13} + 4 \right) } = \lim_{x \to 8} \frac {f(x) - f(8)}{x^{\frac 13} - 2} \cdot \frac 1{x^{\frac 23} + 2x^{\frac 13} + 4}$
what about now?