If `f(x)=lim_(nrarroo)((x^(2)+ax+1)+x^(2n)(2x^(2)+x+b))/(1+x^(2n)) and lim_(xrarrpm1)f(x)` exists, then
The value of b is

Options

A. `-1`
B. 1
C. 0
D. 2

Explanation

To solve the problem, we need to analyze the limit of the function \( f(x) \) as \( n \) approaches infinity and determine the conditions under which the limit exists as \( x \) approaches \( \pm 1 \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \lim_{n \to \infty} \frac{x^2 + ax + 1 + x^{2n}(2x^2 + x + b)}{1 + x^{2n}} \]

2. **Analyze the limit as \( n \to \infty \)**:
- The term \( x^{2n} \) behaves differently depending on the value of \( x \):
- If \( |x| < 1 \), then \( x^{2n} \to 0 \).
- If \( |x| = 1 \), then \( x^{2n} = 1 \).
- If \( |x| > 1 \), then \( x^{2n} \to \infty \).

3. **Evaluate \( f(x) \) for different cases**:
- **Case 1**: \( |x| < 1 \)
\[ f(x) = \frac{x^2 + ax + 1 + 0}{1 + 0} = x^2 + ax + 1 \]
- **Case 2**: \( |x| = 1 \)
\[ f(1) = \frac{1 + a + 1 + 2 + b}{1 + 1} = \frac{3 + a + b}{2} \]
\[ f(-1) = \frac{1 - a + 1 + 2 + b}{1 + 1} = \frac{3 - a + b}{2} \]
- **Case 3**: \( |x| > 1 \)
\[ f(x) = \frac{x^{2n}(2x^2 + x + b)}{x^{2n}} = 2x^2 + x + b \]

4. **Set up conditions for limits at \( x = 1 \) and \( x = -1 \)**:
- For \( x = 1 \):
\[ \lim_{x \to 1^-} f(x) = 1 + a + 1 = 2 + a \]
\[ \lim_{x \to 1^+} f(x) = 2 + 1 + b = 3 + b \]
Setting these equal gives:
\[ 2 + a = 3 + b \quad \text{(1)} \]

- For \( x = -1 \):
\[ \lim_{x \to -1^-} f(x) = 1 - a + 1 = 2 - a \]
\[ \lim_{x \to -1^+} f(x) = 2 + (-1) + b = 1 + b \]
Setting these equal gives:
\[ 2 - a = 1 + b \quad \text{(2)} \]

5. **Solve the system of equations**:
- From equation (1):
\[ a - b = 1 \quad \text{(3)} \]
- From equation (2):
\[ a + b = 1 \quad \text{(4)} \]

6. **Add equations (3) and (4)**:
\[ (a - b) + (a + b) = 1 + 1 \implies 2a = 2 \implies a = 1 \]

7. **Substitute \( a = 1 \) into equation (4)**:
\[ 1 + b = 1 \implies b = 0 \]

### Final Answer:
The value of \( b \) is \( \boxed{0} \).