To evaluate the limit $\lim\limits_{x\to 3}\frac{x^4 -9x^2 + 4x - 12}{(x-3)(4-x)}$, we start by simplifying the expression. Note that $4 - x = -(x - 4)$, so the expression simplifies as follows:

$\frac{x^4 - 9x^2 + 4x - 12}{(x-3)(4-x)} = -\frac{x^4 - 9x^2 + 4x - 12}{(x-3)(x-4)}$

Next, we will attempt to factor the numerator $x^4 - 9x^2 + 4x - 12$. Since we aim to cancel the $(x-3)$ factor in the denominator, we first test whether $x = 3$ is a root of the polynomial. We substitute $x = 3$ into the polynomial:

$3^4 - 9 \cdot 3^2 + 4 \cdot 3 - 12 = 81 - 81 + 12 - 12 = 0$

Since $x = 3$ is a root, $(x-3)$ is a factor of $x^4 - 9x^2 + 4x - 12$.

Similarly, we test whether $x = 4$ is a root:

$4^4 - 9 \cdot 4^2 + 4 \cdot 4 - 12 = 256 - 144 + 16 - 12 = 116 \neq 0$

Thus, $x = 4$ is not a root, meaning there is no $(x-4)$ factor present in the numerator. Thus, we only need to divide $x^4 - 9x^2 + 4x - 12$ by $(x-3)$.

Now, perform polynomial division of $x^4 - 9x^2 + 4x - 12$ by $x-3$:

1. Divide the leading term $x^4$ by the leading term of $x-3$, which gives $x^3$.

2. Multiply $x^3$ by $x-3$ to get $x^4 - 3x^3$.

3. Subtract $x^4 - 3x^3$ from $x^4 + 0x^3 - 9x^2 + 4x - 12$: $x^4 + 0x^3 - 9x^2 + 4x - 12 - (x^4 - 3x^3) = 3x^3 - 9x^2 + 4x - 12$

4. Divide the leading term $3x^3$ by the leading term of $x-3$, resulting in $3x^2$.

5. Multiply $3x^2$ by $x-3$ to get $3x^3 - 9x^2$.

6. Subtract $3x^3 - 9x^2$ from $3x^3 - 9x^2 + 4x - 12$: $3x^3 - 9x^2 + 4x - 12 - (3x^3 - 9x^2) = 4x - 12$

7. Divide $4x$ by $x$ to get 4.

8. Multiply $4$ by $x-3$ to get $4x - 12$.

9. Subtract $4x - 12$ from $4x - 12$ leaves a remainder of 0.

Thus, we have determined that:

$x^4 - 9x^2 + 4x - 12 = (x-3)(x^3 + 3x^2 + 4)$

Now substitute in the expression for the limit:

$-\frac{(x-3)(x^3 + 3x^2 + 4)}{(x-3)(x-4)} = -\frac{x^3 + 3x^2 + 4}{x-4}$

As $x \to 3$, the expression simplifies to substituting $x = 3$:

$-\frac{27 + 27 + 4}{3 - 4} = -\frac{58}{-1} = 58$

Thus, the value of the limit is $\boxed{58}$.