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: Rewrite the expressions using logarithmic properties
We can express \( 2 - \cos(15x) \) and \( \sin(3x) + 1 \) in a form that is easier to analyze as \( x \) approaches 0.

\[ \ln(2 - \cos(15x)) = \ln(1 + (1 - \cos(15x))) \quad \text{and} \quad \ln(\sin(3x) + 1) = \ln(1 + \sin(3x)) \]

### Step 2: Use Taylor series expansions
Using the Taylor series expansions around \( x = 0 \):
- For \( \cos(15x) \):
\[ \cos(15x) \approx 1 - \frac{(15x)^2}{2} = 1 - \frac{225x^2}{2} \]
Thus,
\[ 1 - \cos(15x) \approx \frac{225x^2}{2} \]
So,
\[ \ln(2 - \cos(15x)) \approx \ln\left(1 + \frac{225x^2}{2}\right) \approx \frac{225x^2}{2} \quad \text{(using } \ln(1 + y) \approx y \text{ for small } y\text{)} \]

- For \( \sin(3x) \):
\[ \sin(3x) \approx 3x \]
Thus,
\[ \sin(3x) + 1 \approx 1 + 3x \]
So,
\[ \ln(\sin(3x) + 1) \approx \ln(1 + 3x) \approx 3x \quad \text{(again using the approximation)} \]

### Step 3: Substitute back into the limit
Now substituting these approximations back into the limit:
\[ \lim_{x \to 0} \frac{\ln(2 - \cos(15x))}{\ln^2(\sin(3x) + 1)} \approx \lim_{x \to 0} \frac{\frac{225x^2}{2}}{(3x)^2} \]
Calculating the denominator:
\[ (3x)^2 = 9x^2 \]
Thus, we have:
\[ \lim_{x \to 0} \frac{\frac{225x^2}{2}}{9x^2} = \lim_{x \to 0} \frac{225}{2 \cdot 9} = \frac{225}{18} = 12.5 \]

### Final Answer
The value of the limit is:
\[ \boxed{12.5} \]