Answer
$\frac{4}{5}$
Work Step by Step
Factor each polynomial:
$$\lim_{x\to 1}\frac{x^3-x^2+3x-3}{x^2+3x-4}\\=\lim_{x\to 1}\frac{(x-1)(x^2+3)}{(x-1)(x+4)}.$$
Cancel the common factors: $$\lim_{x\to 1}\frac{(x^2+3)}{(x+4)}\\=\frac{(1^2+3)}{(1+4)}\\=\frac{(1+4)}{5}\\=\frac{4}{5}$$