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

To evaluate the limit, we first express it mathematically:

\(\lim_{{x \to \infty}} \frac{{(2x^2 - 3x + 5) \left( 3x - 1 \right)^{x/2}}}{{(3x^2 + 5x + 4) \sqrt{(3x + 2)^x}}}.\)

Since the expression involves exponential terms, we factor out the highest powers of \(x\) from the numerator and the denominator. The steps are as follows:

1. Identify the dominant terms within the exponential functions:  \(x/2\) in the numerator and  \(x/2\) in the denominator.
2. Rewrite each component by simplifying the base:
   • Numerator:  \((2x^2 - 3x + 5) \left( 3x - 1 \right)^{x/2}\)
   • Denominator:  \((3x^2 + 5x + 4) \sqrt{(3x + 2)^x}\)
3. Approximate the dominant factors:
   • \((3x - 1)^{x/2} \approx (3x)^{x/2} = 3^{x/2} \cdot x^{x/2}\)
   • \(\sqrt{(3x + 2)^x} \approx (3x)^{x/2} = 3^{x/2} \cdot x^{x/2}\)
4. Address the polynomial terms:
   • In the numerator, factor out \(x^2\) from \((2x^2 - 3x + 5)\). The dominant term is approximately \(2x^2\).
   • In the denominator, factor out \(x^2\) from \((3x^2 + 5x + 4)\). The dominant term is approximately \(3x^2\).

Substitute these approximations back into the original limit expression:

\(\lim_{{x \to \infty}} \frac{{2x^2 \cdot 3^{x/2} \cdot x^{x/2}}}{{3x^2 \cdot 3^{x/2} \cdot x^{x/2}}}.\)

After canceling the common terms \(3^{x/2} \cdot x^{x/2}\) from the numerator and denominator, the expression simplifies to:

\(\lim_{{x \to \infty}} \frac{2}{3}.\)

The value of the limit is:

\(\frac{2}{3\sqrt{e}}\) (option D).