Question

lim (m→0) (1 - cos m) m² = ____.

• A. 0
• B. 1
• C. 0.5
• D. 2

Hint:

Alternatively, use L'Hôpital's Rule twice: Differentiate once: $\frac{\sin m}{2m}$ Differentiate again: $\frac{\cos m}{2}$ Substitute $m=0$: $\frac{\cos 0}{2} = \frac{1}{2} = 0.5$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Direct substitution of $m=0$ gives the indeterminate form $0/0$ ($\cos 0 = 1$, so $1-1=0$). We can solve this using trigonometric identities or L'Hôpital's Rule.

Step 2: Key Formula or Approach:
Using the trigonometric identity: \[ 1 - \cos m = 2\sin^2\left(\frac{m}{2}\right) \]

Step 3: Detailed Explanation:
Substitute the identity into the limit: \[ \lim_{m \to 0} \frac{2\sin^2(m/2)}{m^2} \] Rewrite the expression to match the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$: \[ \lim_{m \to 0} 2 \cdot \frac{\sin^2(m/2)}{4 \cdot (m/2)^2} \] \[ = \frac{2}{4} \cdot \lim_{m \to 0} \left[ \frac{\sin(m/2)}{m/2} \right]^2 \] Since the term in the bracket becomes 1: \[ = \frac{1}{2} \cdot (1)^2 = 0.5 \]

Step 4: Final Answer:
The value of the limit is 0.5.