Answer
$$\dfrac{4}{5}$$
Work Step by Step
In order to simplify the given expression, we will use the following rules.
$(a) \lim\limits_{x \to a} \dfrac{p(x)}{q(x)}=\dfrac{\lim\limits_{x \to a} p(x)}{\lim\limits_{x \to a} q(x)} \\ (b) \lim\limits_{x \to a} k(x)=k(a)$ ;
where $a$ is a constant.
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Thus, we have:
$$ \lim\limits_{x \to 1} \dfrac{(x^3-x^2+3x-3)}{(x^2+3x-4)}=\lim\limits_{x \to 1} \dfrac{(x-1)(x^2+3)}{(x-1)(x+4)} \\=\dfrac{\lim\limits_{x \to 1} (x^2+3)}{\lim\limits_{x \to 1} (x +4)}\\=\dfrac{(1)^2+3}{1+4} \\=\dfrac{1+3}{5} \\=\dfrac{4}{5}$$
