$\lim _{x \rightarrow 3} \frac{(84-x)^{\frac{1}{4}}-3}{x-3}$ is

Evaluating $\lim_{x \to 3} \frac{(84-x)^{\frac{1}{4}}-3}{x-3}$ by substitution gives the indeterminate form $\frac{0}{0}$, so we apply L'Hôpital's Rule.

Differentiating numerator and denominator:

$f(x) = (84-x)^{\frac{1}{4}}-3$
$f'(x) = -\frac{1}{4}(84-x)^{-\frac{3}{4}}$

$g(x) = x-3$
$g'(x) = 1$

The limit becomes:

$\lim_{x \to 3} \frac{-\frac{1}{4}(84-x)^{-\frac{3}{4}}}{1} = -\frac{1}{4}(81)^{-\frac{3}{4}}$

Simplifying through exponent rules:

$-\frac{1}{4}(3^4)^{-\frac{3}{4}} = -\frac{1}{4} \cdot 3^{-3} = -\frac{1}{4} \cdot \frac{1}{27} = -\frac{1}{108}$

Final answer: $\boxed{-\frac{1}{108}}$