Question

$$ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = $$

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. 3
• D. 2

Hint:

Memorize $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ as a standard result. It appears frequently as a sub-component in more complex calculus problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a standard trigonometric limit that results in an indeterminate form $0/0$. It can be solved using trigonometric identities or L'Hôpital's Rule.

Step 2: Key Formula or Approach:
Use the identity $1 - \cos x = 2\sin^2(x/2)$ or apply L'Hôpital's Rule twice.

Step 3: Detailed Explanation:
1. Method 1: Trigonometric Identity \[ \lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2} = \lim_{x \to 0} 2 \left( \frac{\sin(x/2)}{x} \right)^2 \] Multiply the denominator by $1/4$ to match the angle: \[ 2 \times \frac{1}{4} \lim_{x \to 0} \left( \frac{\sin(x/2)}{x/2} \right)^2 = \frac{1}{2} (1)^2 = \frac{1}{2} \] 2. Method 2: L'Hôpital's Rule Differentiate numerator and denominator: \[ \lim_{x \to 0} \frac{\sin x}{2x} \] Differentiate again: \[ \lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2} \]

Step 4: Final Answer
The limit is $\frac{1}{2}$.