2017/017) Find positive n such that $lim_{x->3} \frac{x^n-3^n}{x-3} = 108$

we have from definition LHS = $\frac{dy}{dx}$ at x = 3 where $y=x^n$
so differentiating we get $(n)3^{n-1} = 108 = 4 * 3^3$ or n = 4