If $\cot[f(x)] = \dfrac{3x - x^3}{1 - 3x^2}$ and $\sin[g(x)] = \dfrac{1 - x^2}{1 + x^2}$, then $\lim\limits_{x \to t}\dfrac{f(x) - f(t)}{g(x) - g(t)} = \ldots$

Given $\cot[f(x)] = \dfrac{3x - x^3}{1 - 3x^2}$

Substituting $x = \tan\theta$, we get:

$\cot[f(x)] = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan(3\theta)$

$\cot[f(x)] = \cot\left(\dfrac{\pi}{2} - 3\theta\right)$

$\Rightarrow f(x) = \dfrac{\pi}{2} - 3\theta = \dfrac{\pi}{2} - 3\tan^{-1}x$

Differentiating with respect to $x$, we get:

$f'(x) = \dfrac{-3}{1 + x^2}$

Similarly, given $\sin[g(x)] = \dfrac{1 - x^2}{1 + x^2}$

Substituting $x = \tan\theta$, we get:

$\sin[g(x)] = \dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos(2\theta)$

$\sin[g(x)] = \sin\left(\dfrac{\pi}{2} - 2\theta\right)$

$\Rightarrow g(x) = \dfrac{\pi}{2} - 2\theta = \dfrac{\pi}{2} - 2\tan^{-1}x$

Differentiating with respect to $x$, we get:

$g'(x) = \dfrac{-2}{1 + x^2}$

The given limit is in the form of the definition of the derivative (or can be evaluated using L'Hospital's Rule):

$\lim\limits_{x \to t}\dfrac{f(x) - f(t)}{g(x) - g(t)} = \dfrac{f'(t)}{g'(t)}$

Substituting the derivatives at $x = t$:

$\dfrac{f'(t)}{g'(t)} = \dfrac{\dfrac{-3}{1 + t^2}}{\dfrac{-2}{1 + t^2}} = \dfrac{3}{2}$

Answer: $\dfrac{3}{2}$