The value of `lim_(x to 0) (15^(x) - 5^(x) - 3^(x) + 1)/(1 - cos 2x) ` is

Options

A. `((log 3) (log 5))/(2)`
B. 2 (log 3) (log 5)
C. `(log 3 + log 5)/(2)`
D. None of these

Explanation

To solve the limit

\[ \lim_{x \to 0} \frac{15^x - 5^x - 3^x + 1}{1 - \cos 2x} \]

we will follow these steps:

### Step 1: Analyze the numerator and denominator

As \( x \to 0 \), both the numerator and denominator approach 0. Thus, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and denominator.

### Step 2: Differentiate the numerator and denominator

1. **Numerator**: Differentiate \( 15^x - 5^x - 3^x + 1 \):
\[ \frac{d}{dx}(15^x) = 15^x \ln(15), \quad \frac{d}{dx}(5^x) = 5^x \ln(5), \quad \frac{d}{dx}(3^x) = 3^x \ln(3) \]
Therefore, the derivative of the numerator is:
\[ 15^x \ln(15) - 5^x \ln(5) - 3^x \ln(3) \]

2. **Denominator**: Differentiate \( 1 - \cos(2x) \):
\[ \frac{d}{dx}(1 - \cos(2x)) = 2 \sin(2x) \]

### Step 3: Rewrite the limit using derivatives

Now we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{15^x \ln(15) - 5^x \ln(5) - 3^x \ln(3)}{2 \sin(2x)} \]

### Step 4: Evaluate the limit as \( x \to 0 \)

Substituting \( x = 0 \) into the derivatives:
- \( 15^0 = 1 \), \( 5^0 = 1 \), \( 3^0 = 1 \)
- Thus, the numerator becomes:
\[ 1 \cdot \ln(15) - 1 \cdot \ln(5) - 1 \cdot \ln(3) = \ln(15) - \ln(5) - \ln(3) \]
Using the properties of logarithms:
\[ \ln(15) - \ln(5) - \ln(3) = \ln\left(\frac{15}{5 \cdot 3}\right) = \ln(1) = 0 \]

The denominator becomes:
\[ 2 \sin(0) = 0 \]

Since we still have an indeterminate form \( \frac{0}{0} \), we apply L'Hôpital's Rule again.

### Step 5: Differentiate again

1. **Numerator**: Differentiate again:
\[ \frac{d}{dx}(15^x \ln(15) - 5^x \ln(5) - 3^x \ln(3)) = 15^x (\ln(15))^2 - 5^x (\ln(5))^2 - 3^x (\ln(3))^2 \]

2. **Denominator**: Differentiate again:
\[ \frac{d}{dx}(2 \sin(2x)) = 4 \cos(2x) \]

### Step 6: Rewrite the limit again

Now we have:
\[ \lim_{x \to 0} \frac{15^x (\ln(15))^2 - 5^x (\ln(5))^2 - 3^x (\ln(3))^2}{4 \cos(2x)} \]

### Step 7: Substitute \( x = 0 \)

Substituting \( x = 0 \):
- The numerator becomes:
\[ 1 \cdot (\ln(15))^2 - 1 \cdot (\ln(5))^2 - 1 \cdot (\ln(3))^2 = (\ln(15))^2 - (\ln(5))^2 - (\ln(3))^2 \]
- The denominator becomes:
\[ 4 \cos(0) = 4 \]

### Step 8: Final limit evaluation

Thus, the limit is:
\[ \frac{(\ln(15))^2 - (\ln(5))^2 - (\ln(3))^2}{4} \]

### Conclusion

The final value of the limit is:
\[ \frac{1}{2} \left( \ln(3) \cdot \ln(5) \right) \]