Question

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

• A. \( a = 1, b = 2 \)
• B. \( a = 2, b = 1 \)
• C. \( a = 1, b \in \mathbb{R} \)
• D. None of these

Hint:

When dealing with limits of the form \( \left( 1 + \frac{1}{n} \right)^n \), use the approximation \( \left( 1 + \frac{a}{x} \right)^x \approx e^a \) for large \( x \).

Answer 1

The Correct Option is C

Step 1: Recognize the limit form.
We are given the limit:
\[ \lim_{x \to \infty} \left( 1 + \frac{a}{x} + \frac{b}{x^2} \right)^{2x} = e^2 \] This is a standard limit form that resembles the limit definition of \( e \). The expression inside the limit looks like a binomial expansion of \( \left( 1 + \frac{1}{n} \right)^n \).

Step 2: Use the approximation for large \( x \).
For large values of \( x \), \( \frac{a}{x} \) and \( \frac{b}{x^2} \) become very small. So, we can approximate the expression as: \[ \left( 1 + \frac{a}{x} + \frac{b}{x^2} \right)^{2x} \approx \left( 1 + \frac{a}{x} \right)^{2x} \]

Step 3: Apply the binomial approximation.
We recognize that: \[ \lim_{x \to \infty} \left( 1 + \frac{a}{x} \right)^{2x} = e^{2a} \] Thus, the limit becomes: \[ e^{2a} = e^2 \]

Step 4: Solve for \( a \).
From \( e^{2a} = e^2 \), we find \( 2a = 2 \), so \( a = 1 \).

Step 5: Conclusion.
Thus, \( a = 1 \), and \( b \) can be any real number. Therefore, the correct answer is option (C), \( a = 1, b \in \mathbb{R} \).