The value of `lim_(xrarr0)(ln(10-9cos2x))/(ln^(2)(sin3x+1))` is equal to

Answer

To solve the limit \( \lim_{x \to 0} \frac{\ln(10 - 9 \cos 2x)}{\ln^2(\sin 3x + 1)} \), we will follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the limit in a more manageable form:
\[ \lim_{x \to 0} \frac{\ln(10 - 9 \cos 2x)}{\ln^2(\sin 3x + 1)} \]

### Step 2: Expand the logarithm in the numerator
Using the fact that \( \cos 2x \) approaches 1 as \( x \) approaches 0, we can use the Taylor expansion:
\[ \cos 2x \approx 1 - \frac{(2x)^2}{2} = 1 - 2x^2 \]
Thus,
\[ 10 - 9 \cos 2x \approx 10 - 9(1 - 2x^2) = 10 - 9 + 18x^2 = 1 + 18x^2 \]
Now we can rewrite the logarithm:
\[ \ln(10 - 9 \cos 2x) \approx \ln(1 + 18x^2) \approx 18x^2 \quad \text{(using } \ln(1 + u) \approx u \text{ for small } u\text{)} \]

### Step 3: Expand the logarithm in the denominator
For the denominator, we can use a similar expansion:
\[ \sin 3x \approx 3x \quad \text{(as } x \to 0\text{)} \]
Thus,
\[ \sin 3x + 1 \approx 3x + 1 \]
And then,
\[ \ln(\sin 3x + 1) \approx \ln(1 + 3x) \approx 3x \quad \text{(again using } \ln(1 + u) \approx u\text{)} \]
So,
\[ \ln^2(\sin 3x + 1) \approx (3x)^2 = 9x^2 \]

### Step 4: Substitute back into the limit
Now we substitute back into the limit:
\[ \lim_{x \to 0} \frac{18x^2}{9x^2} = \lim_{x \to 0} 2 = 2 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{2} \]