If the graph of the function `y=f(x)` has a unique tangent at point `(e^(a),0)` (not vertical, non-horizontal) on the graph, then evaluate `lim_(xtoe^(a))log_e((1 +9f(x))-"tan"(f(x)))/(2f(x))`.

Answer

To solve the limit problem, we will follow these steps:

### Step 1: Identify the given information
We know that the function \( y = f(x) \) has a unique tangent at the point \( (e^a, 0) \). This means:
- \( f(e^a) = 0 \)
- \( f'(e^a) \) exists and is not equal to zero (since the tangent is neither vertical nor horizontal).

### Step 2: Set up the limit expression
We need to evaluate the limit:
\[ \lim_{x \to e^a} \frac{\log(1 + 9f(x)) - \tan(f(x))}{2f(x)} \]

### Step 3: Substitute \( x = e^a \)
Substituting \( x = e^a \) into the limit gives:
- \( f(e^a) = 0 \)
- Thus, \( 9f(e^a) = 0 \)
- \( \log(1 + 9f(e^a)) = \log(1) = 0 \)
- \( \tan(f(e^a)) = \tan(0) = 0 \)

This results in an indeterminate form \( \frac{0 - 0}{0} = \frac{0}{0} \).

### Step 4: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:
\[ \lim_{x \to e^a} \frac{\frac{d}{dx}[\log(1 + 9f(x)) - \tan(f(x))]}{\frac{d}{dx}[2f(x)]} \]

### Step 5: Differentiate the numerator and denominator
1. **Differentiate the numerator**:
- The derivative of \( \log(1 + 9f(x)) \) is:
\[ \frac{9f'(x)}{1 + 9f(x)} \]
- The derivative of \( \tan(f(x)) \) is:
\[ \sec^2(f(x)) \cdot f'(x) \]
- Therefore, the derivative of the numerator is:
\[ \frac{9f'(x)}{1 + 9f(x)} - \sec^2(f(x)) f'(x) \]

2. **Differentiate the denominator**:
- The derivative of \( 2f(x) \) is:
\[ 2f'(x) \]

### Step 6: Rewrite the limit using derivatives
Now we have:
\[ \lim_{x \to e^a} \frac{\frac{9f'(x)}{1 + 9f(x)} - \sec^2(f(x)) f'(x)}{2f'(x)} \]

### Step 7: Factor out \( f'(x) \)
Assuming \( f'(x) \neq 0 \) near \( e^a \), we can factor \( f'(x) \):
\[ = \lim_{x \to e^a} \frac{f'(x) \left( \frac{9}{1 + 9f(x)} - \sec^2(f(x)) \right)}{2f'(x)} \]
This simplifies to:
\[ = \lim_{x \to e^a} \frac{9}{1 + 9f(x)} - \sec^2(f(x)) \cdot \frac{1}{2} \]

### Step 8: Substitute \( f(e^a) = 0 \)
Substituting \( f(e^a) = 0 \):
\[ = \frac{9}{1 + 9 \cdot 0} - \sec^2(0) \cdot \frac{1}{2} \]
\[ = \frac{9}{1} - 1 \cdot \frac{1}{2} \]
\[ = 9 - \frac{1}{2} = \frac{18}{2} - \frac{1}{2} = \frac{17}{2} \]

### Step 9: Final answer
Thus, the limit evaluates to:
\[ \frac{17}{2} \]