Question

The value of \(\lim_{x \to +\infty}\sqrt[3]{\frac{3x+5}{6x-8}}\) is given by:

• A. $\sqrt[3]{\frac{1}{3}}$
• B. $\sqrt[3]{\frac{1}{5}}$
• C. $\sqrt[3]{\frac{1}{7}}$
• D. $\sqrt[3]{\frac{1}{2}}$

Hint:

For limits at infinity of rational functions where the degree of the numerator equals the degree of the denominator, the limit of the fraction is simply the ratio of their leading coefficients: $\frac{3}{6} = \frac{1}{2}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To evaluate limits of algebraic fractions as $x \to \infty$, we divide the numerator and the denominator by the highest power of $x$ present in the expression.
Detailed Explanation:
Let us evaluate the limit: \[ L = \lim_{x \to +\infty} \sqrt[3]{\frac{3x+5}{6x-8}} \] Since the cube root function is continuous, we can move the limit inside the radical: \[ L = \sqrt[3]{\lim_{x \to +\infty} \frac{3x+5}{6x-8}} \] Divide every term in the numerator and the denominator by $x$: \[ L = \sqrt[3]{\lim_{x \to +\infty} \frac{3 + \frac{5}{x}}{6 - \frac{8}{x}}} \] As $x \to +\infty$, the terms $\frac{5}{x}$ and $\frac{8}{x}$ approach zero: \[ L = \sqrt[3]{\frac{3 + 0}{6 - 0}} \] \[ L = \sqrt[3]{\frac{3}{6}} = \sqrt[3]{\frac{1}{2}} \] Thus, the limit value is $\sqrt[3]{\frac{1}{2}}$.

Step 2: Final Answer:
The value is $\sqrt[3]{\frac{1}{2}}$, which corresponds to Option (D).