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:

For limits involving exponential functions, be familiar with the standard limit \( \lim_{x \to \infty} \left( 1 + \frac{a}{x} \right)^{x} = e^a \), which can simplify many problems involving exponential growth.

Answer 1

The Correct Option is C

We start with the given limit: \[ \lim_{x \to \infty} \left( 1 + \frac{a}{x} + \frac{b}{x^2} \right)^{2x} = e^2 \] We can rewrite this as: \[ \lim_{x \to \infty} \left( 1 + \frac{a}{x} \right)^{2x} \times \lim_{x \to \infty} \left( 1 + \frac{b}{x^2} \right)^{2x} \] The first limit is well known: \[ \lim_{x \to \infty} \left( 1 + \frac{a}{x} \right)^{2x} = e^{2a} \] For the second term, since \( b/x^2 \to 0 \) as \( x \to \infty \), we have: \[ \lim_{x \to \infty} \left( 1 + \frac{b}{x^2} \right)^{2x} = 1 \] Therefore, the limit is: \[ e^{2a} \times 1 = e^2 \] Thus, \( 2a = 2 \), so \( a = 1 \). The value of \( b \) can be any real number because it does not affect the limit.