Question

The value of $ \lim_{x \to \infty} \left( \frac{x^2 - 2x + 1}{x^2 - 4x + 2} \right)^{2x} \text{ is} $

• A. \( e^2 \)
• B. \( e^4 \)
• C. \( e \)
• D. \( e^{16} \)

Hint:

When evaluating limits of the form \( \left( 1 + \frac{a}{x} \right)^{bx} \) as \( x \to \infty \), recognize that it converges to \( e^{ab} \).

Answer 1

The Correct Option is B

We need to find the value of the limit \[ L = \lim_{x \to \infty} \left( \frac{x^2 - 2x + 1}{x^2 - 4x + 2} \right)^{2x}. \]
Step 1: Simplify the expression inside the limit
For large values of \( x \), the dominant terms in the numerator and denominator are \( x^2 \). Hence, the expression can be approximated as: \[ \frac{x^2 - 2x + 1}{x^2 - 4x + 2} \approx \frac{x^2}{x^2} = 1. \] However, we need to refine this approximation to capture the behavior for large \( x \).
Step 2: Refine the approximation
Factor the numerator and denominator to focus on the behavior of smaller terms: \[ \frac{x^2 - 2x + 1}{x^2 - 4x + 2} = \frac{1 - \frac{2}{x} + \frac{1}{x^2}}{1 - \frac{4}{x} + \frac{2}{x^2}}. \] As \( x \to \infty \), the higher-order terms vanish, so we get: \[ \frac{1 - \frac{2}{x}}{1 - \frac{4}{x}}. \] Now, expand the expression using the approximation \( \frac{1}{1 - z} \approx 1 + z \) for small \( z \): \[ \frac{1 - \frac{2}{x}}{1 - \frac{4}{x}} \approx 1 + \frac{2}{x}. \]
Step 3: Take the limit
Now, substitute this back into the original expression for the limit: \[ L = \lim_{x \to \infty} \left( 1 + \frac{2}{x} \right)^{2x}. \] This is a standard limit, and we know that: \[ \lim_{x \to \infty} \left( 1 + \frac{a}{x} \right)^x = e^a. \] Thus, we get: \[ L = e^4. \] Hence, the value of the limit is \( e^4 \).