Question

Evaluate: $$ \lim_{x \to \infty} \left( \sqrt[3]{x^3 + 4x^2} - \sqrt{x^2 - 3x} \right) $$

• A. \(\frac{17}{6}\)
• B. \(\frac{25}{6}\)
• C. \(-\frac{1}{6}\)
• D. \(\frac{37}{6}\)

Hint:

Use binomial expansion for large \( x \) to simplify expressions.

Answer 1

The Correct Option is A

Rewrite terms: \[ \sqrt[3]{x^3 + 4x^2} = \sqrt[3]{x^3\left(1 + \frac{4}{x}\right)} = x \sqrt[3]{1 + \frac{4}{x}} \] \[ \sqrt{x^2 - 3x} = x \sqrt{1 - \frac{3}{x}} \] Use binomial expansion: \[ \sqrt[3]{1 + \frac{4}{x}} \approx 1 + \frac{4}{3x} - \dots \] \[ \sqrt{1 - \frac{3}{x}} \approx 1 - \frac{3}{2x} - \dots \] Therefore: \[ \sqrt[3]{x^3 + 4x^2} - \sqrt{x^2 - 3x} \approx x \left(1 + \frac{4}{3x}\right) - x \left(1 - \frac{3}{2x}\right) = \frac{4}{3} + \frac{3}{2} = \frac{8}{6} + \frac{9}{6} = \frac{17}{6} \]