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.

Answer 2

The question asks for \( \displaystyle\lim_{m \to 0} \frac{1 - \cos m}{m^2} \), which gives the indeterminate form \( \frac{0}{0} \) on direct substitution. Since both numerator and denominator are differentiable, we can resolve this using L'Hopital's Rule instead of a trigonometric identity, and check why each of the other listed values would arise from a plausible slip.

1. 0: This value would result from wrongly concluding the whole limit is \( 0/0 = 0 \) without applying L'Hopital's Rule or any series expansion, treating the indeterminate form as if it simply evaluated to zero. That is not valid.
2. 1: Applying L'Hopital's Rule once, differentiate numerator and denominator with respect to \( m \), gives \( \displaystyle\lim_{m\to 0}\frac{\sin m}{2m} \), which is still \( 0/0 \). If one stops here and incorrectly treats \( \frac{\sin m}{m} \) as equal to \( 1 \) while forgetting the factor of \( 2 \) in the denominator, one would wrongly get \( 1 \) instead of \( \frac{1}{2} \).
3. 0.5: Applying L'Hopital's Rule a second time on \( \displaystyle\lim_{m\to 0}\frac{\sin m}{2m} \): differentiate again to get \( \displaystyle\lim_{m\to 0}\frac{\cos m}{2} \). Substituting \( m = 0 \) gives \( \frac{\cos 0}{2} = \frac{1}{2} = 0.5 \). This is the correctly resolved limit.
4. 2: This value would result from inverting the fraction at some stage, for instance mistakenly computing \( \displaystyle\lim_{m\to0}\frac{2m}{\sin m} \) instead of \( \displaystyle\lim_{m\to0}\frac{\sin m}{2m} \), which flips the correct answer to a value larger than 1 instead of less than 1.

Applying L'Hopital's Rule twice consistently gives \( \frac{1}{2} \), confirming the value obtained by other methods as well.

Therefore, the correct answer is 0.5.