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:
Show Hint
- \( \frac{2}{\sqrt{3e}} \)
- \( \frac{2e}{\sqrt{3}} \)
- \( \frac{2e}{3} \)
- \( \frac{2}{3\sqrt{e}} \)
The Correct Option is D
Approach Solution - 1
To evaluate the given limit, we first express it mathematically:
\(\lim_{{x \to \infty}} \frac{{(2x^2 - 3x + 5) \left( 3x - 1 \right)^{x/2}}}{{(3x^2 + 5x + 4) \sqrt{(3x + 2)^x}}}.\)
The expression involves exponential terms, and it is beneficial to factor out the highest powers of \(x\) from both the numerator and the denominator. Let's perform the simplification step-by-step:
- Notice that the dominant term inside the exponential functions for both numerator and denominator is \(x/2\) or \(x/2\), respectively.
- Rewriting each component with base simplification:
- For the numerator: \((2x^2 - 3x + 5) \left( 3x - 1 \right)^{x/2}\)
- For the denominator: \((3x^2 + 5x + 4) \sqrt{(3x + 2)^x}\)
- We start with approximating the dominant factors:
- \((3x - 1)^{x/2} \approx (3x)^{x/2} = 3^{x/2} \cdot x^{x/2}\)
- \(\sqrt{(3x + 2)^x} \approx (3x)^{x/2} = 3^{x/2} \cdot x^{x/2}\)
- For the polynomial terms:
- In the numerator: factor out \(x^2\) from \((2x^2 - 3x + 5)\). Approximately, \(2x^2\) is the dominant term.
- In the denominator: factor out \(x^2\) from \((3x^2 + 5x + 4)\). Approximately, \(3x^2\) is the dominant term.
Substitute these back into the original limit:
\(\lim_{{x \to \infty}} \frac{{2x^2 \cdot 3^{x/2} \cdot x^{x/2}}}{{3x^2 \cdot 3^{x/2} \cdot x^{x/2}}}.\)
Canceling out the common terms \(3^{x/2} \cdot x^{x/2}\) from both the numerator and the denominator, we simplify the expression to:
\(\lim_{{x \to \infty}} \frac{2}{3}.\)
Therefore, the value of the limit is:
\(\frac{2}{3\sqrt{e}}\) (option D).
Approach Solution -2
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