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 can follow these steps:

### Step 1: Rewrite the logarithmic expressions
We can use the fact that \( \ln(1 + x) \approx x \) as \( x \to 0 \). Thus, we rewrite the logarithmic expressions:
\[ \ln(2 - \cos(25x)) = \ln(1 + (1 - \cos(25x))) \approx 1 - \cos(25x) \]
and
\[ \ln(\sin(3x) + 1) = \ln(1 + \sin(3x)) \approx \sin(3x) \]

### Step 2: Substitute the approximations into the limit
Substituting these approximations into the limit gives:
\[ \lim_{x \to 0} \frac{9(1 - \cos(25x))}{5(\sin(3x))^2} \]

### Step 3: Use trigonometric limits
We know that:
\[ 1 - \cos(x) \approx \frac{x^2}{2} \quad \text{as } x \to 0 \]
Thus,
\[ 1 - \cos(25x) \approx \frac{(25x)^2}{2} = \frac{625x^2}{2} \]
And for \( \sin(3x) \):
\[ \sin(3x) \approx 3x \quad \text{as } x \to 0 \]
So,
\[ (\sin(3x))^2 \approx (3x)^2 = 9x^2 \]

### Step 4: Substitute back into the limit
Now substituting these approximations back into the limit:
\[ \lim_{x \to 0} \frac{9 \cdot \frac{625x^2}{2}}{5 \cdot 9x^2} = \lim_{x \to 0} \frac{9 \cdot 625x^2/2}{45x^2} \]

### Step 5: Simplify the limit
The \( x^2 \) terms cancel out:
\[ = \lim_{x \to 0} \frac{9 \cdot 625/2}{45} = \frac{5625/2}{45} = \frac{5625}{90} = 62.5 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{62.5} \]