If `f:RrarrR` is a continuous function satisfying `f(0)=1` and `f(2x)-f(x)=xAAxepsilonR` and `lim_(nrarroo)(f(x)-f(x/(2^(n))))=P(x)`. Then `P(x)` is

Options

A. a constant function
B. a linear function
C. a quadratic polynomial in `x`
D. a cubic polynomial in `x`

Explanation

To solve the given problem step by step, we will analyze the function \( f \) and the conditions provided.

### Step 1: Understanding the Function
We are given that \( f: \mathbb{R} \to \mathbb{R} \) is a continuous function satisfying:
- \( f(0) = 1 \)
- \( f(2x) - f(x) = x \) for all \( x \in \mathbb{R} \)

### Step 2: Define the Function \( g(x) \)
Let's define a new function:
\[ g(x) = f(2x) - f(x) \]
From the given condition, we know:
\[ g(x) = x \]

### Step 3: Evaluate \( g\left(\frac{x}{2}\right) \)
Now, let's evaluate \( g\left(\frac{x}{2}\right) \):
\[ g\left(\frac{x}{2}\right) = f(x) - f\left(\frac{x}{2}\right) = \frac{x}{2} \]

### Step 4: Evaluate \( g\left(\frac{x}{4}\right) \)
Next, we evaluate \( g\left(\frac{x}{4}\right) \):
\[ g\left(\frac{x}{4}\right) = f\left(\frac{x}{2}\right) - f\left(\frac{x}{4}\right) = \frac{x}{4} \]

### Step 5: Generalize for \( g\left(\frac{x}{2^n}\right) \)
Continuing this process, we can generalize:
\[ g\left(\frac{x}{2^n}\right) = f\left(\frac{x}{2^{n-1}}\right) - f\left(\frac{x}{2^n}\right) = \frac{x}{2^n} \]

### Step 6: Summing Up the Function Values
Now we can sum these equations:
\[ g(x) + g\left(\frac{x}{2}\right) + g\left(\frac{x}{4}\right) + \ldots + g\left(\frac{x}{2^n}\right) = x + \frac{x}{2} + \frac{x}{4} + \ldots + \frac{x}{2^n} \]

The left-hand side simplifies to:
\[ f(2x) - f\left(\frac{x}{2^n}\right) \]

The right-hand side is a geometric series:
\[ x \left(1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^n}\right) = x \cdot \left(\frac{1 - \left(\frac{1}{2}\right)^{n+1}}{1 - \frac{1}{2}}\right) = x \cdot (2 - \frac{x}{2^n}) \]

### Step 7: Taking the Limit as \( n \to \infty \)
Taking the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} \left(f(2x) - f\left(\frac{x}{2^n}\right)\right) = \lim_{n \to \infty} \left(x \cdot 2\right) = 2x \]
Since \( f\left(\frac{x}{2^n}\right) \) approaches \( f(0) = 1 \):
\[ f(2x) - 1 = 2x \implies f(2x) = 2x + 1 \]

### Step 8: Finding \( P(x) \)
Now we need to find:
\[ \lim_{n \to \infty} \left(f(x) - f\left(\frac{x}{2^n}\right)\right) \]
Using the previous results, we have:
\[ f(x) - 1 = 2x \implies P(x) = 2x \]

### Conclusion
Thus, the function \( P(x) \) is:
\[ P(x) = 2x \]