Question

Evaluate the limit: \[ \lim_{x \to \infty} \left( \frac{3x^2 - 2x + 3}{3x^2 + x - 2} \right)^{3x - 2} \]

• A. \( -3 \)
• B. \( e^{-1} \)
• C. \( e^{-3} \)
• D. \( -1 \)

Hint:

For limits in the form \( (1^\infty) \), take the logarithm and apply first-order approximations. Use the dominant terms in numerator and denominator for large \( x \).

Answer 1

The Correct Option is C

Step 1: Simplifying the fraction inside the limit Divide both numerator and denominator by \( x^2 \): \[ \frac{3x^2 - 2x + 3}{3x^2 + x - 2} = \frac{3 - \frac{2}{x} + \frac{3}{x^2}}{3 + \frac{1}{x} - \frac{2}{x^2}} \] As \( x \to \infty \), the terms \( \frac{2}{x} \), \( \frac{3}{x^2} \), \( \frac{1}{x} \), and \( \frac{2}{x^2} \) tend to zero. Thus, \[ \frac{3x^2 - 2x + 3}{3x^2 + x - 2} \to \frac{3}{3} = 1. \] Step 2: Applying Logarithm for Exponential Limit Form We have an indeterminate form \( (1^\infty) \), so we take logarithms: \[ L = \lim_{x \to \infty} (3x - 2) \ln \left( \frac{3x^2 - 2x + 3}{3x^2 + x - 2} \right). \] Expanding using first-order approximations: \[ \frac{3x^2 - 2x + 3}{3x^2 + x - 2} = 1 + \frac{-3x - 5}{3x^2 + x - 2}. \] Approximating for large \( x \), \[ \ln \left( 1 + \frac{-3x - 5}{3x^2 + x - 2} \right) \approx \frac{-3x - 5}{3x^2 + x - 2}. \] Step 3: Evaluating the Limit Multiplying by \( (3x - 2) \), \[ L = \lim_{x \to \infty} (3x - 2) \cdot \frac{-3x - 5}{3x^2 + x - 2}. \] Approximating, \[ L = \lim_{x \to \infty} \frac{(3x - 2)(-3x - 5)}{3x^2 + x - 2}. \] For large \( x \), the highest degree term dominates: \[ L = \lim_{x \to \infty} \frac{-9x^2 - 15x + 6x + 10}{3x^2 + x - 2}. \] \[ = \lim_{x \to \infty} \frac{-9x^2 - 9x + 10}{3x^2 + x - 2}. \] Dividing by \( x^2 \), \[ = \lim_{x \to \infty} \frac{-9 - \frac{9}{x} + \frac{10}{x^2}}{3 + \frac{1}{x} - \frac{2}{x^2}}. \] For large \( x \), \[ = \frac{-9}{3} = -3. \] Thus, \[ L = -3. \] Exponentiating both sides, \[ \lim_{x \to \infty} \left( \frac{3x^2 - 2x + 3}{3x^2 + x - 2} \right)^{3x - 2} = e^{-3}. \]