Step 1: Understanding the Concept:
When we substitute $x = 3$ directly into the expression, we get $\frac{(84-3)^{1/4} - 3}{3-3} = \frac{81^{1/4} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}$. This is an indeterminate form. To resolve this, we can apply L'Hôpital's Rule or use an algebraic substitution to simplify the limit into a standard recognizable form.
Step 2: Key Formula or Approach:
Approach 1: L'Hôpital's Rule: If $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0}$, then it equals $\lim_{x \to a} \frac{f'(x)}{g'(x)}$.
Approach 2: Standard Limit Formula: $\lim_{t \to a} \frac{t^n - a^n}{t - a} = n a^{n-1}$ after an appropriate substitution.
Step 3: Detailed Explanation:
Let's use L'Hôpital's Rule, which is straightforward here.
Let $f(x) = (84-x)^{\frac{1}{4}} - 3$ and $g(x) = x - 3$.
Differentiate the numerator with respect to $x$:
\[ f'(x) = \frac{d}{dx} \left[ (84-x)^{\frac{1}{4}} - 3 \right] = \frac{1}{4}(84-x)^{\frac{1}{4}-1} \cdot \frac{d}{dx}(84-x) \]
\[ f'(x) = \frac{1}{4}(84-x)^{-\frac{3}{4}} \cdot (-1) = -\frac{1}{4}(84-x)^{-\frac{3}{4}} \]
Differentiate the denominator with respect to $x$:
\[ g'(x) = \frac{d}{dx} [x - 3] = 1 \]
Apply L'Hôpital's Rule:
\[ \lim_{x \to 3} \frac{f(x)}{g(x)} = \lim_{x \to 3} \frac{-\frac{1}{4}(84-x)^{-\frac{3}{4}}}{1} \]
Now, substitute $x = 3$ directly into the new expression:
\[ = -\frac{1}{4}(84-3)^{-\frac{3}{4}} = -\frac{1}{4}(81)^{-\frac{3}{4}} \]
Recognize that $81 = 3^4$:
\[ = -\frac{1}{4}(3^4)^{-\frac{3}{4}} \]
Using the exponent rule $(a^m)^n = a^{m \cdot n}$:
\[ = -\frac{1}{4}(3^{4 \cdot (-\frac{3}{4})}) = -\frac{1}{4}(3^{-3}) \]
\[ = -\frac{1}{4} \cdot \frac{1}{3^3} = -\frac{1}{4} \cdot \frac{1}{27} = -\frac{1}{108} \]
Step 4: Final Answer:
The limit evaluates to $\frac{-1}{108}$.