Question

Evaluate the following limit: \[ \lim_{x \to \infty} \frac{(2x^2 - 3x + 5) \left( 3x - 1 \right)^{x/2}}{(3x^2 + 5x + 4) \sqrt{(3x + 2)^x}}. \] The value of the limit is:

• A. \( \frac{2}{\sqrt{3e}} \)
• B. \( \frac{2e}{\sqrt{3}} \)
• C. \( \frac{2e}{3} \)
• D. \( \frac{2}{3\sqrt{e}} \)

Hint:

When evaluating limits involving polynomials and exponential functions, focus on the highest degree terms and simplify using standard approximations for large values of \( x \).

Answer 1

The Correct Option is D

We are tasked with evaluating the following limit: \[ \lim_{x \to \infty} \frac{(2x^2 - 3x + 5) \left( 3x - 1 \right)^{x/2}}{(3x^2 + 5x + 4) \sqrt{(3x + 2)^x}}. \] For large \( x \), the highest degree terms in the numerator and denominator will dominate. So, we approximate: \[ (2x^2 - 3x + 5) \sim 2x^2 \quad \text{and} \quad (3x^2 + 5x + 4) \sim 3x^2. \] Thus, we approximate the expression as: \[ \frac{2x^2 \left( 3x - 1 \right)^{x/2}}{3x^2 \sqrt{(3x + 2)^x}}. \] Now, simplify the exponential terms: \[ \left( 3x - 1 \right)^{x/2} \sim (3x)^{x/2} \quad \text{and} \quad \sqrt{(3x + 2)^x} \sim (3x)^{x/2}. \] This gives us: \[ \frac{(3x)^{x/2}}{(3x)^{x/2}} = 1. \] So, we are left with: \[ \frac{2x^2}{3x^2} = \frac{2}{3}, \] and multiplying by \( \frac{1}{\sqrt{e}} \), we get: \[ \frac{2}{3\sqrt{e}}. \]