Step 1: Understanding the Concept:
As $x \to 1$, the base $\log_3 3x \to \log_3 3 = 1$ and the exponent $\log_x 8 \to \infty$.
This is a $1^\infty$ indeterminate form. Step 2: Key Formula or Approach:
For $\lim f(x)^{g(x)}$ of type $1^\infty$, result is $e^{\lim (f(x)-1)g(x)}$.
Using $\log_a b = \frac{\ln b}{\ln a}$. Step 3: Detailed Explanation:
$f(x) = \log_3 3x = \log_3 3 + \log_3 x = 1 + \log_3 x$.
Then $f(x) - 1 = \log_3 x$.
The limit becomes $e^L$ where:
\[ L = \lim_{x \to 1} (\log_3 x)(\log_x 8) = \lim_{x \to 1} \frac{\ln x}{\ln 3} \cdot \frac{\ln 8}{\ln x} = \frac{\ln 8}{\ln 3} = \log_3 8 \]
So the limit is $e^{\log_3 8}$. Step 4: Final Answer:
The correct value is $e^{\log_3 8}$.