Question

$\lim_{x \to 1} (\log_3 3x)^{\log_x 8} = .......$

• A. $\text{e}^{\log_3 8}$
• B. $\log_8 3$
• C. $e^{\log_8 3}$
• D. $\log_3 8$

Hint:

For $1^\infty$ forms, use $\lim f(x)^{g(x)} = e^{\lim g(x)[f(x)-1]}$.

Answer 1

The Correct Option is A

Step 1: Simplify the base
Base $= \log_3 (3x) = \log_3 3 + \log_3 x = 1 + \log_3 x$.
Step 2: Form $1^\infty$
As $x \to 1$, base $\to 1$ and exponent $\to \infty$.
Limit $= e^{\lim_{x \to 1} (\text{base} - 1) \cdot \text{exponent}}$
$= e^{\lim_{x \to 1} (\log_3 x) \cdot \frac{\log 8}{\log x}}$
Step 3: Calculation
Using change of base: $\log_3 x = \frac{\log x}{\log 3}$.
Exponent $= \frac{\log x}{\log 3} \cdot \frac{\log 8}{\log x} = \frac{\log 8}{\log 3} = \log_3 8$.
Result $= e^{\log_3 8}$.
Final Answer: (A)