If `lim_(xtooo)(1+a/x+b/(x^(2)))^(2x)=e^2` then values of a and b are

Options

A. `a=1,b=2`
B. `a=2,b=1`
C. `a=1,b epsilon R`
D. None of these

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. **Rewrite the Limit**:
We start by rewriting the limit expression:
\[ \lim_{x \to \infty} \left(1 + \frac{a}{x} + \frac{b}{x^2}\right)^{2x} \]

2. **Identify the Form**:
As \(x\) approaches infinity, \(\frac{a}{x}\) approaches 0 and \(\frac{b}{x^2}\) also approaches 0. Thus, we can express the limit in the form of \(1 + u\) where \(u\) approaches 0:
\[ 1 + \frac{a}{x} + \frac{b}{x^2} \approx 1 + \frac{a}{x} \]
for large \(x\).

3. **Use the Exponential Limit**:
We can use the limit property:
\[ \lim_{x \to \infty} \left(1 + \frac{u}{n}\right)^{n} = e^u \]
Here, we can set \(u = 2a\) and \(n = x\). Thus, we rewrite our limit:
\[ \lim_{x \to \infty} \left(1 + \frac{a}{x} + \frac{b}{x^2}\right)^{2x} = \lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^{2x} \cdot \left(1 + \frac{b}{x^2}\right)^{2x} \]

4. **Evaluate Each Part**:
The first part:
\[ \lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^{2x} = e^{2a} \]
The second part:
\[ \lim_{x \to \infty} \left(1 + \frac{b}{x^2}\right)^{2x} \to 1 \text{ (since } \frac{b}{x^2} \to 0 \text{)} \]

5. **Combine the Results**:
Therefore, we have:
\[ \lim_{x \to \infty} \left(1 + \frac{a}{x} + \frac{b}{x^2}\right)^{2x} = e^{2a} \]

6. **Set the Equation**:
Setting this equal to \(e^2\):
\[ e^{2a} = e^2 \]

7. **Solve for \(a\)**:
Since the bases are the same, we can equate the exponents:
\[ 2a = 2 \implies a = 1 \]

8. **Determine \(b\)**:
From the limit evaluation, we see that the term involving \(b\) does not affect the limit as \(x\) approaches infinity. Thus, \(b\) can be any real number.

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