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: Expand \( \cos m \) as a Taylor series around \( m = 0 \): \[ \cos m = 1 - \frac{m^2}{2} + \frac{m^4}{24} - \cdots \]

Step 2: Substitute this into the numerator: \[ 1 - \cos m = \frac{m^2}{2} - \frac{m^4}{24} + \cdots \]

Step 3: Divide by \( m^2 \) and take the limit as \( m \to 0 \), so every term with a positive power of \( m \) vanishes: \[ \lim_{m \to 0} \frac{1-\cos m}{m^2} = \lim_{m \to 0} \left( \frac{1}{2} - \frac{m^2}{24} + \cdots \right) = \frac{1}{2} \] \[ \boxed{0.5} \]