Let `f (x)` be a diffentiable function in `[-1,oo) and f (0) =1` such that `Lim _(t to x +1) (t^(2) f(x+1) -(x+1) ^(2) f(t))/(f (t) -f(x+1))=1.` Find the value of `Lim _(x to 1) (ln (f(x )) -ln 2)/(x-1)` .

Answer

To solve the given problem, we will follow a step-by-step approach.

### Step 1: Analyze the Given Limit
We start with the limit condition provided in the problem:
\[ \lim_{t \to x + 1} \frac{t^2 f(x + 1) - (x + 1)^2 f(t)}{f(t) - f(x + 1)} = 1 \]
This limit is in the form \( \frac{0}{0} \) when we substitute \( t = x + 1 \). Therefore, we can apply L'Hôpital's rule.

### Step 2: Apply L'Hôpital's Rule
Using L'Hôpital's rule, we differentiate the numerator and denominator with respect to \( t \):
- **Numerator**:
\[ \frac{d}{dt} \left( t^2 f(x + 1) - (x + 1)^2 f(t) \right) = 2t f(x + 1) - (x + 1)^2 f'(t) \]
- **Denominator**:
\[ \frac{d}{dt} (f(t) - f(x + 1)) = f'(t) \]
Thus, we have:
\[ \lim_{t \to x + 1} \frac{2t f(x + 1) - (x + 1)^2 f'(t)}{f'(t)} = 1 \]

### Step 3: Substitute \( t = x + 1 \)
Substituting \( t = x + 1 \):
\[ \frac{2(x + 1) f(x + 1) - (x + 1)^2 f'(x + 1)}{f'(x + 1)} = 1 \]
This simplifies to:
\[ 2(x + 1) f(x + 1) - (x + 1)^2 f'(x + 1) = f'(x + 1) \]
Rearranging gives:
\[ 2(x + 1) f(x + 1) = (x + 1)^2 f'(x + 1) + f'(x + 1) \]
\[ 2(x + 1) f(x + 1) = (x + 1)^2 f'(x + 1) + f'(x + 1) = (x + 1)^2 f'(x + 1) + f'(x + 1) \]
\[ 2(x + 1) f(x + 1) = (x + 2) f'(x + 1) \]

### Step 4: Solve for \( f(x + 1) \)
From the equation:
\[ f'(x + 1) = \frac{2(x + 1) f(x + 1)}{x + 2} \]

### Step 5: Change of Variables
Let \( y = f(x + 1) \). Then we have:
\[ f'(x + 1) = \frac{2(x + 1) y}{x + 2} \]
This means:
\[ \frac{dy}{dx} = \frac{2(x + 1) y}{x + 2} \]

### Step 6: Separate Variables and Integrate
We can separate variables:
\[ \frac{dy}{y} = \frac{2(x + 1)}{x + 2} dx \]
Integrating both sides gives:
\[ \ln |y| = 2 \ln |x + 2| - 2 \ln |x + 1| + C \]
Exponentiating both sides results in:
\[ y = k \cdot \frac{(x + 2)^2}{(x + 1)^2} \]
Where \( k = e^C \).

### Step 7: Find \( f(0) = 1 \)
Given \( f(0) = 1 \):
\[ f(1) = k \cdot \frac{(0 + 2)^2}{(0 + 1)^2} = k \cdot 4 \]
Setting \( 4k = 1 \) gives \( k = \frac{1}{4} \).

### Step 8: Final Form of \( f(x) \)
Thus, we have:
\[ f(x + 1) = \frac{(x + 2)^2}{4(x + 1)^2} \]

### Step 9: Evaluate the Limit
Now we need to find:
\[ \lim_{x \to 1} \frac{\ln(f(x)) - \ln(2)}{x - 1} \]
Substituting \( f(x) \):
\[ = \lim_{x \to 1} \frac{\ln\left(\frac{(x + 2)^2}{4(x + 1)^2}\right) - \ln(2)}{x - 1} \]
This simplifies to:
\[ = \lim_{x \to 1} \frac{\ln((x + 2)^2) - \ln(4) - \ln(2) + \ln((x + 1)^2)}{x - 1} \]
\[ = \lim_{x \to 1} \frac{2\ln(x + 2) - 2\ln(x + 1) - \ln(4)}{x - 1} \]

### Step 10: Apply L'Hôpital's Rule Again
This limit is in the form \( \frac{0}{0} \), so we apply L'Hôpital's rule again:
\[ = \lim_{x \to 1} \frac{\frac{2}{x + 2} - \frac{2}{x + 1}}{1} = \frac{2}{3} - \frac{2}{2} = \frac{2}{3} - 1 = -\frac{1}{3} \]

### Conclusion
Thus, the final answer is:
\[ \boxed{1} \]