#### Answer

$$\dfrac{13}{28}$$

#### 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:
$$ \lim\limits_{x \to 3} \dfrac{(x^3-3x^2+4x-12)}{(x^4-3x^3+x-3)}=\lim\limits_{x \to 3} \dfrac{(x-3)(x^2+4)}{(x-3)(x^3+1)} \\=\dfrac{\lim\limits_{x \to 3} (x^2+4)}{\lim\limits_{x \to 3} (x^3 +1)}\\=\dfrac{3^2+4}{3^3+1} \\=\dfrac{9+4}{27+1} \\=\dfrac{13}{28}$$