The value of `lim_(xrarr0)(ln(2-cos15x))/(ln^(2)(sin3x+1))` is equal to

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{\ln(2 - \cos(15x))}{\ln^2(\sin(3x) + 1)}, \]

we will follow these steps:

### Step 1: Simplify the numerator
We can use the fact that as \( x \to 0 \), \( \cos(15x) \to 1 \). Thus, we can rewrite the numerator:

\[ \ln(2 - \cos(15x)) = \ln(2(1 - \frac{\cos(15x)}{2})). \]

Using the Taylor series expansion for \( \cos(15x) \) around \( x = 0 \):

\[ \cos(15x) \approx 1 - \frac{(15x)^2}{2} = 1 - \frac{225x^2}{2}. \]

So,

\[ 2 - \cos(15x) \approx 2 - \left(1 - \frac{225x^2}{2}\right) = 1 + \frac{225x^2}{2}. \]

Thus,

\[ \ln(2 - \cos(15x)) \approx \ln\left(1 + \frac{225x^2}{2}\right). \]

### Step 2: Apply the logarithmic limit
Using the limit property \( \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1 \), we can express:

\[ \ln\left(1 + \frac{225x^2}{2}\right) \approx \frac{225x^2}{2} \text{ as } x \to 0. \]

### Step 3: Simplify the denominator
Now, consider the denominator:

\[ \ln^2(\sin(3x) + 1). \]

As \( x \to 0 \), \( \sin(3x) \approx 3x \), so:

\[ \sin(3x) + 1 \approx 3x + 1. \]

Thus,

\[ \ln(\sin(3x) + 1) \approx \ln(1 + 3x) \approx 3x \text{ as } x \to 0. \]

So,

\[ \ln^2(\sin(3x) + 1) \approx (3x)^2 = 9x^2. \]

### Step 4: Substitute back into the limit
Now, substituting back into the limit:

\[ \lim_{x \to 0} \frac{\frac{225x^2}{2}}{(9x^2)} = \lim_{x \to 0} \frac{225}{2 \cdot 9} = \frac{225}{18} = 12.5. \]

### Final Answer
Thus, the value of the limit is:

\[ \boxed{12.5}. \]