Question

The value of \(\lim_{x \to \infty} \frac{x}{x + \frac{\sqrt[3]{x}}{x + \frac{\sqrt[3]{x}}{x + \dots}}}\) is

• A. 0
• B. 1
• C. 2
• D. 1/2

Hint:

Set up a recursive equation \(y = f(y)\) and take the limit.

Answer 1

The Correct Option is B

To solve the problem, we need to find the value of the following limit:

\(\lim_{x \to \infty} \frac{x}{x + \frac{\sqrt[3]{x}}{x + \frac{\sqrt[3]{x}}{x + \dots}}}\).

Let's simplify and analyze the given expression:

1. The denominator of the fraction inside the given limit is a recursive sequence: \(x + \frac{\sqrt[3]{x}}{\ldots}\).
2. Let \(y = x + \frac{\sqrt[3]{x}}{y}\).
3. Multiply both sides by \(y\) to eliminate the fraction:
4. \(y^2 = xy + \sqrt[3]{x}\).
5. Rearrange the equation:
6. \(y^2 - xy - \sqrt[3]{x} = 0\).
7. This is a quadratic equation in \(y\). Applying the quadratic formula:
8. \(y = \frac{x \pm \sqrt{x^2 + 4\sqrt[3]{x}}}{2}\).
9. For large \(x\), the term \(4\sqrt[3]{x}\) becomes negligible compared to \(x^2\), so we approximate:
10. \(y \approx \frac{x \pm x}{2}\).
11. Since we are interested in the positive value (as it makes physical sense in real-world applications):
12. \(y \approx x\).
13. Now substitute \(y\) back into the original limit expression:
14. The expression simplifies to:
15. \(\lim_{x \to \infty} \frac{x}{y} = \lim_{x \to \infty} \frac{x}{x}\).
16. This limit clearly evaluates to:
17. \(1\).

Therefore, the value of the limit is \(1\). This matches the given correct answer option.

Conclusion: The value of \(\lim_{x \to \infty} \frac{x}{x + \frac{\sqrt[3]{x}}{x + \frac{\sqrt[3]{x}}{x + \dots}}}\) is indeed \(1\), which confirms that the correct option is 1.