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: Write the Maclaurin series expansion of \(\sin m\) near \(m=0\): \(\sin m = m - \dfrac{m^3}{6} + \dfrac{m^5}{120} - \cdots\).

Step 2: Divide every term by \(m\): \(\dfrac{\sin m}{m} = 1 - \dfrac{m^2}{6} + \dfrac{m^4}{120} - \cdots\).

Step 3: As \(m \to 0\), every term beyond the first contains a positive power of \(m\) and vanishes, leaving only the constant term.
\[ \lim_{m \to 0} \frac{\sin m}{m} = \boxed{1} \]