`lim_(x to 9)[log_(2) log_(3)x]^(1/(x-9)`=

Options

A. 1
B. 0
C. `infty`
D. `-infty`

Answer

To solve the limit \( \lim_{x \to 9} \left[ \log_2 (\log_3 x) \right]^{\frac{1}{x-9}} \), we will follow these steps:

### Step 1: Identify the Form
First, we substitute \( x = 9 \) into the expression to see what form we get:
\[ \log_3 9 = \log_3 (3^2) = 2 \quad \text{(since } 9 = 3^2\text{)} \]
Then,
\[ \log_2 (\log_3 9) = \log_2 2 = 1 \]
Thus, we have:
\[ \left[ \log_2 (\log_3 x) \right]^{\frac{1}{x-9}} \to 1^{\frac{1}{0}} \text{ as } x \to 9 \]
This is an indeterminate form of type \( 1^\infty \).

### Step 2: Apply the Exponential Limit
We can rewrite the limit using the exponential function:
\[ L = \lim_{x \to 9} \left[ \log_2 (\log_3 x) \right]^{\frac{1}{x-9}} = e^{\lim_{x \to 9} \left( \log_2 (\log_3 x) - 1 \right) \cdot \frac{1}{x-9}} \]

### Step 3: Simplify the Limit
Now we need to evaluate:
\[ \lim_{x \to 9} \left( \log_2 (\log_3 x) - 1 \right) \cdot \frac{1}{x-9} \]
We know that \( \log_2 (\log_3 x) \to 1 \) as \( x \to 9 \), so we can apply L'Hôpital's Rule since we have a \( 0/0 \) form.

### Step 4: Differentiate the Numerator and Denominator
Let \( f(x) = \log_2 (\log_3 x) - 1 \). We need to differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( \log_2 (\log_3 x) \right) = \frac{1}{\log 2} \cdot \frac{1}{\log 3} \cdot \frac{1}{x} \]
Thus,
\[ \lim_{x \to 9} \frac{f(x)}{x-9} = \lim_{x \to 9} \frac{f'(x)}{1} = \frac{1}{\log 2 \cdot \log 3} \cdot \frac{1}{9} \]

### Step 5: Calculate the Final Limit
Now we substitute \( x = 9 \):
\[ \lim_{x \to 9} \frac{1}{\log 2 \cdot \log 3 \cdot 9} \]
Thus,
\[ L = e^{\frac{1}{9 \log 2 \cdot \log 3}} \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 9} \left[ \log_2 (\log_3 x) \right]^{\frac{1}{x-9}} = e^{\frac{1}{9 \log 2 \cdot \log 3}} \]

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