Question

lim (m→0) 5sinm m =

• A. 0
• B. 1
• C. ∞
• D. not defined

Hint:

This limit only holds true when the angle $m$ is measured in radians. If the angle were in degrees, the limit would be $\pi/180$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem uses the fundamental trigonometric limit property which states that as the angle approaches zero, the ratio of the sine of the angle to the angle itself tends to unity.

Step 2: Key Formula or Approach:
The standard limit is: \[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

Step 3: Detailed Explanation:
The expression is: \[ \lim_{m \to 0} \frac{5 \sin m}{m} \] By the constant multiple rule of limits: \[ 5 \times \left( \lim_{m \to 0} \frac{\sin m}{m} \right) \] Applying the standard limit: \[ 5 \times 1 = 5 \] Correction Note: Based on the provided options, if "1" is intended as the answer, the "5" in the numerator was likely a typo in the original question paper. If "5" is not present, the answer is 1. Given the strict options, (B) 1 is the closest conceptual answer for the base identity.

Step 4: Final Answer:
The value of the limit is 5. (If the question was intended as $\lim \frac{\sin m}{m}$, the answer is 1).

Answer 2

The expression asks for \( \displaystyle\lim_{m \to 0} \frac{\sin m}{m} \), which takes the indeterminate form \(\frac{0}{0}\) when \(m\) is substituted directly, since \(\sin(0)=0\). L'Hopital's Rule applies directly to this form: differentiate the numerator and denominator separately and then take the limit of that new ratio.

\[ \lim_{m \to 0} \frac{\sin m}{m} = \lim_{m \to 0} \frac{\frac{d}{dm}(\sin m)}{\frac{d}{dm}(m)} = \lim_{m \to 0} \frac{\cos m}{1} = \cos(0) = 1 \]

1. 0: This would be the result only if the numerator vanished faster than the denominator near \(m=0\), but differentiating shows the ratio settles at \(\cos(0)=1\), not 0.
2. 1: This matches exactly what L'Hopital's Rule produces: differentiating \(\sin m\) gives \(\cos m\), and \(\cos(0)=1\).
3. \(\infty\): This would only occur if the denominator vanished while the numerator stayed finite and nonzero, but here both numerator and denominator vanish together at the same rate, giving a finite limit, not an infinite one.
4. Not defined: The limit is not undefined, since L'Hopital's Rule is valid precisely because both the numerator and denominator are differentiable near \(m=0\) and the denominator's derivative is nonzero there.

The differentiation-based check confirms the ratio approaches a clean, finite value as \(m\) shrinks to zero.

Therefore, the correct answer is 1.