Compute the following limits using L'Hospital's rule if appropriate. Use INF to denote $\infty$ and MINF to denote $-\infty$.

$\displaystyle \lim_{x\to 1} \frac{3^x-3}{x^2-1}$ =

$\displaystyle \lim_{x\to \infty} \frac{\tan^{-1}(x)}{(1/x) - 3}$ =