Question

Evaluate the limit: \[ \lim\limits_{x \to 6} \frac{\sqrt{x^2 + 13} - 7}{x^2 - 36}. \]

• A. \( \frac{1}{7} \)
• B. \( \frac{1}{13} \)
• C. \( \frac{13}{36} \)
• D. \( \frac{1}{14} \)
• E. \( \frac{1}{36} \)

Hint:

When a limit results in a \( \frac{0}{0} \) form, apply L'Hôpital’s Rule by differentiating the numerator and denominator.

Answer 1

The Correct Option is D

Step 1: Substitute \( x = 6 \) Substituting directly: \[ \sqrt{6^2 + 13} - 7 = \sqrt{36 + 13} - 7 = \sqrt{49} - 7 = 7 - 7 = 0. \] \[ x^2 - 36 = 6^2 - 36 = 36 - 36 = 0. \] Since both numerator and denominator become zero, apply L'Hôpital's Rule. 
Step 2: Differentiate numerator and denominator Differentiate the numerator: \[ \frac{d}{dx} \left( \sqrt{x^2 + 13} - 7 \right) = \frac{1}{2} \cdot \frac{2x}{\sqrt{x^2 + 13}} = \frac{x}{\sqrt{x^2 + 13}}. \] Differentiate the denominator: \[ \frac{d}{dx} (x^2 - 36) = 2x. \] 
Step 3: Compute the limit \[ \lim\limits_{x \to 6} \frac{\frac{x}{\sqrt{x^2 + 13}}}{2x} = \lim\limits_{x \to 6} \frac{1}{2 \sqrt{x^2 + 13}}. \] Substituting \( x = 6 \): \[ \frac{1}{2 \sqrt{36 + 13}} = \frac{1}{2 \times 7} = \frac{1}{14}. \] Thus, the correct answer is: \[ \frac{1}{14}. \]