If `lim_(x -> oo) (1 + a/x + b/x^2)^(2x)= e^2` then the values of a and b, are

Options

A. `a epsilonR, bepsilonR`
B. `a=1, b epsilonR`
C. `a epsilonR, b=2`
D. `a=1, b=2`

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to \infty} \left(1 + \frac{a}{x} + \frac{b}{x^2}\right)^{2x} = e^2 \]

### Step-by-Step Solution:

1. **Identify the form of the limit**: As \( x \to \infty \), both \( \frac{a}{x} \) and \( \frac{b}{x^2} \) approach 0. Therefore, the expression inside the limit approaches \( 1 \). This gives us an indeterminate form \( 1^{\infty} \).

2. **Use the exponential limit transformation**: For limits of the form \( 1^{\infty} \), we can rewrite it using the exponential function:
\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} (f(x) - 1) g(x)} \]
Here, \( f(x) = 1 + \frac{a}{x} + \frac{b}{x^2} \) and \( g(x) = 2x \).

3. **Calculate \( f(x) - 1 \)**:
\[ f(x) - 1 = \frac{a}{x} + \frac{b}{x^2} \]

4. **Multiply by \( g(x) \)**:
\[ (f(x) - 1) g(x) = \left(\frac{a}{x} + \frac{b}{x^2}\right) \cdot 2x = 2a + \frac{2b}{x} \]

5. **Take the limit as \( x \to \infty \)**:
\[ \lim_{x \to \infty} \left(2a + \frac{2b}{x}\right) = 2a + 0 = 2a \]

6. **Set the limit equal to the given value**:
Since we know from the problem statement that:
\[ e^{2a} = e^2 \]
We can equate the exponents:
\[ 2a = 2 \]

7. **Solve for \( a \)**:
\[ a = 1 \]

8. **Determine the value of \( b \)**: Since \( \frac{2b}{x} \) approaches 0 as \( x \to \infty \), \( b \) can be any real number. Thus, \( b \) is not constrained and can take any value in the real numbers.

### Final Values:
- \( a = 1 \)
- \( b \) can be any real number.