Question

\[ \mathop {\lim }\limits_{x \to 1}\left(\log _3 3 x\right)^{\log _x 8}=\ldots \]

• A. \(\mathrm{e}^{\mathrm{log}_{3}8}\)
• B. \(\mathrm{log}_{8}3\)
• C. \(e^{\mathrm{log}_{8}3}\)
• D. \(\mathrm{log}_{3}8\)

Hint:

The limit has the form 1 to the power infinity; use e raised to the limit of (base - 1) times the exponent.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
As \(x\to1\), \(\log_{3}3x=1+\log_{3}x\to1\) and \(\log_{x}8=\dfrac{\log 8}{\log x}\to\infty\). This is the form \(1^{\infty}\).

Step 2: Use the rule for 1 to the infinity
\(\lim f^{g}=e^{\lim g(f-1)}\).

Step 3: Compute the exponent
\(g(f-1)=\dfrac{\log 8}{\log x}\cdot\log_{3}x=\dfrac{\log8}{\log x}\cdot\dfrac{\log x}{\log3}=\log_{3}8\).

Step 4: Result
The limit is \(e^{\log_{3}8}\).

Final Answer:
The limit is e raised to log base 3 of 8. \[ \boxed{\text{(A) }e^{\log_{3}8}} \]