If `"lt"_(x to infty)[1 +a/x + b/x^(2)]^(2x)= e^(2)`, then (a,b) is:

Options

A. (2,1)
B. (1,2)
C. [1, any real constant]
D. (1,1)

Answer

To solve the limit problem, we start with the expression given in the question:

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

### Step 1: Rewrite the limit expression
As \( x \to \infty \), both \( \frac{a}{x} \) and \( \frac{b}{x^2} \) approach 0. Thus, we can rewrite the limit expression:

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

### Step 2: Apply the logarithm
To simplify the limit, we can take the natural logarithm of both sides:

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

Using properties of logarithms, we can move the limit inside:

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

### Step 3: Expand the logarithm using Taylor series
For small values of \( y \), \( \ln(1+y) \approx y \). Thus, we can approximate:

\[ \ln\left(1 + \frac{a}{x} + \frac{b}{x^2}\right) \approx \frac{a}{x} + \frac{b}{x^2} \]

### Step 4: Substitute the approximation into the limit
Substituting this approximation into our limit gives:

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

As \( x \to \infty \), the term \( \frac{2b}{x} \) approaches 0:

\[ \lim_{x \to \infty} \left(2a + 0\right) = 2a \]

### Step 5: Set the limit equal to 2
Now we set this equal to the right-hand side of our earlier equation:

\[ 2a = 2 \]

### Step 6: Solve for \( a \)
Dividing both sides by 2 gives:

\[ a = 1 \]

### Step 7: Determine \( b \)
Since the term \( \frac{2b}{x} \) approaches 0, \( b \) can be any real constant. Therefore, \( b \) is not restricted and can take any value.

### Final Answer
Thus, the values of \( (a, b) \) are:

\[ (a, b) = (1, \text{any real constant}) \]