The value of `lim_(xrarr0)(9ln (2-cos 25x))/(5ln^(2)(sin 3x+1))` is equal to

Answer

To solve the limit \( \lim_{x \to 0} \frac{9 \ln(2 - \cos(25x))}{5 \ln^2(\sin(3x) + 1)} \), we will follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the limit in a more manageable form. We recognize that as \( x \to 0 \), both the numerator and denominator approach 0, creating a \( \frac{0}{0} \) indeterminate form.

\[ \lim_{x \to 0} \frac{9 \ln(2 - \cos(25x))}{5 \ln^2(\sin(3x) + 1)} \]

### Step 2: Use the approximation for \( \cos \)
Using the Taylor series expansion for \( \cos \) around 0, we have:

\[ \cos(25x) \approx 1 - \frac{(25x)^2}{2} = 1 - \frac{625x^2}{2} \]

Thus,

\[ 2 - \cos(25x) \approx 2 - \left(1 - \frac{625x^2}{2}\right) = 1 + \frac{625x^2}{2} \]

### Step 3: Apply the logarithm approximation
Using the approximation \( \ln(1 + u) \approx u \) for small \( u \):

\[ \ln(2 - \cos(25x)) \approx \ln\left(1 + \frac{625x^2}{2}\right) \approx \frac{625x^2}{2} \]

### Step 4: Substitute into the limit
Substituting this back into our limit gives:

\[ \lim_{x \to 0} \frac{9 \cdot \frac{625x^2}{2}}{5 \ln^2(\sin(3x) + 1)} \]

### Step 5: Approximate \( \sin(3x) \)
Using the Taylor series for \( \sin(3x) \):

\[ \sin(3x) \approx 3x \]

Thus,

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

### Step 6: Apply the logarithm approximation again
Using the logarithm approximation again:

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

### Step 7: Substitute back into the limit
Now substituting this into the limit gives:

\[ \lim_{x \to 0} \frac{9 \cdot \frac{625x^2}{2}}{5 (3x)^2} \]

### Step 8: Simplify the expression
This simplifies to:

\[ \lim_{x \to 0} \frac{9 \cdot \frac{625x^2}{2}}{5 \cdot 9x^2} = \lim_{x \to 0} \frac{625}{10} = 62.5 \]

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

\[ \boxed{62.5} \]