Question

Evaluate the limit: \[ \lim_{x \to \infty} \frac{\sqrt{\cos^2 x + 3} - \sqrt{\cos^2 x + \sin x + 3}}{x} \]

• A. \( 0 \)
• B. \( 1 \)
• C. \( -1 \)
• D. Undefined

Hint:

When evaluating limits involving oscillating functions like sine or cosine, focus on the behavior of the denominator, especially if it involves \( x \) in the denominator. If the denominator grows without bound, the limit will approach zero.

Answer 1

The Correct Option is A

Evaluate the limit: \[ L = \lim_{x \to \infty} \frac{\sqrt{\cos^2 x + 3} - \sqrt{\cos^2 x + \sin x + 3}}{x} \]

1. Step 1: Simplify the expression: As \( x \to \infty \), the numerator involves square roots of trigonometric expressions. We will start by simplifying the numerator using the conjugate: \[ \text{Numerator} = \sqrt{\cos^2 x + 3} - \sqrt{\cos^2 x + \sin x + 3} \] Multiply by the conjugate of the numerator: \[ \text{Conjugate} = \sqrt{\cos^2 x + 3} + \sqrt{\cos^2 x + \sin x + 3} \] The numerator becomes: \[ \frac{(\cos^2 x + 3) - (\cos^2 x + \sin x + 3)}{x \left( \sqrt{\cos^2 x + 3} + \sqrt{\cos^2 x + \sin x + 3} \right)} \] Simplifying the numerator: \[ = \frac{-\sin x}{x \left( \sqrt{\cos^2 x + 3} + \sqrt{\cos^2 x + \sin x + 3} \right)} \]

2. Step 2: Analyze the behavior as \( x \to \infty \): As \( x \to \infty \), the term \( \sin x \) oscillates between \( -1 \) and \( 1 \), so the numerator remains bounded. The denominator, however, grows without bound due to the factor of \( x \). Therefore, the entire expression approaches \( 0 \).

3. Step 3: Conclusion: Hence, the limit is: \[ L = 0 \]