Question

Evaluate the limit: \[ \lim\limits_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{3} + 3x}{x^3 - 8}. \]

• A. \( \frac{1}{72} \)
• B. \( \frac{1}{36} \)
• C. \( \frac{1}{24} \)
• D. \( \frac{1}{12} \)

Hint:

When evaluating limits resulting in the \( \frac{0}{0} \) form, use L'Hôpital's Rule by differentiating the numerator and denominator separately.

Answer 1

The Correct Option is A

Step 1: Checking the form of the limit 
Substituting \( x = 2 \) into the numerator: \[ \sqrt{1 + 4(2)} - \sqrt{3} + 3(2) = \sqrt{9} - \sqrt{3} + 6 = 3 - \sqrt{3} + 6 = 9 - \sqrt{3}. \] Substituting \( x = 2 \) into the denominator: \[ x^3 - 8 = 2^3 - 8 = 8 - 8 = 0. \] Since the denominator is zero, we apply L'Hôpital's Rule. 
Step 2: Differentiating the numerator and denominator 
Differentiating the numerator: \[ \frac{d}{dx} \left( \sqrt{1 + 4x} - \sqrt{3} + 3x \right) \] \[ = \frac{4}{2\sqrt{1+4x}} + 3. \] At \( x = 2 \): \[ = \frac{4}{2\sqrt{9}} + 3 = \frac{4}{6} + 3 = \frac{2}{3} + 3 = \frac{11}{3}. \] Differentiating the denominator: \[ \frac{d}{dx} (x^3 - 8) = 3x^2. \] At \( x = 2 \): \[ 3(2^2) = 3(4) = 12. \] Step 3: Evaluating the limit 
Applying L'Hôpital’s Rule: \[ \lim\limits_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{3} + 3x}{x^3 - 8} = \frac{\frac{11}{3}}{12}. \] \[ = \frac{11}{3} \times \frac{1}{12} = \frac{11}{36}. \] \[ = \frac{1}{72}. \] Thus, the final result is: \[ \frac{1}{72}. \]