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 a is

Options

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

Explanation

To solve the problem, we need to analyze the function given and find the value of \( a \) such that 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. **Consider the cases for \( x \)**:
- **Case 1**: When \( |x| < 1 \) (i.e., \( -1 < x < 1 \))
- **Case 2**: When \( |x| > 1 \) (i.e., \( x > 1 \) or \( x < -1 \))

3. **Evaluate \( f(x) \) for \( |x| < 1 \)**:
- As \( n \to \infty \), \( x^{2n} \to 0 \) because \( |x| < 1 \).
- Therefore, the function simplifies to:
\[ f(x) = \frac{x^2 + ax + 1 + 0}{1 + 0} = x^2 + ax + 1 \]

4. **Evaluate \( f(x) \) for \( |x| > 1 \)**:
- As \( n \to \infty \), \( x^{2n} \to \infty \) because \( |x| > 1 \).
- The function simplifies to:
\[ f(x) = \frac{0 + x^{2n}(2x^2 + x + b)}{x^{2n}} = 2x^2 + x + b \]

5. **Set up the limits at \( x = -1 \)**:
- From the left (approaching from \( |x| < 1 \)):
\[ \lim_{x \to -1^-} f(x) = (-1)^2 + a(-1) + 1 = 1 - a + 1 = 2 - a \]
- From the right (approaching from \( |x| > 1 \)):
\[ \lim_{x \to -1^+} f(x) = 2(-1)^2 + (-1) + b = 2 - 1 + b = 1 + b \]

6. **Set the limits equal for continuity**:
\[ 2 - a = 1 + b \quad \text{(1)} \]

7. **Set up the limits at \( x = 1 \)**:
- From the left (approaching from \( |x| < 1 \)):
\[ \lim_{x \to 1^-} f(x) = 1^2 + a(1) + 1 = 1 + a + 1 = 2 + a \]
- From the right (approaching from \( |x| > 1 \)):
\[ \lim_{x \to 1^+} f(x) = 2(1)^2 + (1) + b = 2 + 1 + b = 3 + b \]

8. **Set the limits equal for continuity**:
\[ 2 + a = 3 + b \quad \text{(2)} \]

9. **Solve the system of equations**:
- From equation (1):
\[ b = 2 - a - 1 = 1 - a \]
- Substitute \( b \) into equation (2):
\[ 2 + a = 3 + (1 - a) \]
\[ 2 + a = 4 - a \]
\[ 2a = 2 \implies a = 1 \]

10. **Find \( b \)**:
- Substitute \( a = 1 \) into \( b = 1 - a \):
\[ b = 1 - 1 = 0 \]

### Conclusion:
The value of \( a \) is \( \boxed{1} \).