If a function satisfies the relation `f(x) f''(x)-f(x)f'(x)=(f'(x))^(2) AA x in R and f(0)=f'(0)=1,` then
The value of `lim_(x to -oo) f(x)` is

Options

A. `(1)/(sqrt(e ))`
B. `(1)/(e )`
C. `sqrt(e )`
D. e

Explanation

To solve the problem, we start with the given relation:

\[ f(x) f''(x) - f(x) f'(x) = (f'(x))^2 \]

We also know that:

\[ f(0) = 1 \quad \text{and} \quad f'(0) = 1 \]

### Step 1: Rearranging the given equation

We can rearrange the given equation as follows:

\[ f''(x) - \frac{f'(x)}{f(x)} f'(x) = \frac{(f'(x))^2}{f(x)} \]

This simplifies to:

\[ f''(x) - f'(x) = \frac{(f'(x))^2}{f(x)} \]

### Step 2: Separating variables

We can separate the variables by rewriting the equation:

\[ \frac{f''(x)}{f'(x)} - 1 = \frac{f'(x)}{f(x)} \]

### Step 3: Integrating both sides

Now, we integrate both sides with respect to \(x\):

\[ \int \left( \frac{f''(x)}{f'(x)} - 1 \right) dx = \int \frac{f'(x)}{f(x)} dx \]

The left side becomes:

\[ \ln |f'(x)| - x \]

And the right side becomes:

\[ \ln |f(x)| \]

Thus, we have:

\[ \ln |f'(x)| - x = \ln |f(x)| + C \]

### Step 4: Exponentiating both sides

Exponentiating both sides gives us:

\[ \frac{|f'(x)|}{|f(x)|} = e^{x + C} \]

We can rewrite this as:

\[ |f'(x)| = K |f(x)| e^x \]

where \(K = e^C\).

### Step 5: Solving the differential equation

Now we have a separable differential equation:

\[ \frac{f'(x)}{f(x)} = K e^x \]

Integrating both sides gives us:

\[ \ln |f(x)| = K e^x + C_1 \]

Exponentiating again leads to:

\[ f(x) = A e^{K e^x} \]

where \(A = e^{C_1}\).

### Step 6: Applying initial conditions

Using the initial condition \(f(0) = 1\):

\[ f(0) = A e^{K} = 1 \implies A = e^{-K} \]

Thus, we have:

\[ f(x) = e^{-K} e^{K e^x} = e^{K(e^x - 1)} \]

### Step 7: Finding the limit as \(x \to -\infty\)

Now, we need to find:

\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} e^{K(e^x - 1)} \]

As \(x \to -\infty\), \(e^x \to 0\), so:

\[ K(e^x - 1) \to K(0 - 1) = -K \]

Thus:

\[ f(x) \to e^{-K} \]

### Conclusion

Since \(K\) is a constant, we can conclude that:

\[ \lim_{x \to -\infty} f(x) = e^{-K} \]

Given the initial conditions, we find that \(K = 1\), so:

\[ \lim_{x \to -\infty} f(x) = e^{-1} = \frac{1}{e} \]

### Final Answer:

\[ \lim_{x \to -\infty} f(x) = \frac{1}{e} \]