​$$\lim_{\theta \to 0} \frac{1 - \cos{m\theta}}{1 - \cos{n\theta}} = \, ?$$​​

Correct option is B

​$$\text{For small values of } \theta, \text{ we use the approximation for } \cos(x) \text{ when } x \text{ is small:} \\\cos(x) \approx 1 - \frac{x^2}{2} \\\text{Thus, for small } \theta, \text{ we can approximate:} \\1 - \cos(m\theta) \approx \frac{(m\theta)^2}{2} = m^2 \frac{\theta^2}{2} \\\text{and} \\1 - \cos(n\theta) \approx \frac{(n\theta)^2}{2} = n^2 \frac{\theta^2}{2} \\\textbf{Substitute these approximations into the limit} \\\text{Substituting the approximations into the given limit expression:} \\\lim_{\theta \to 0} \frac{1 - \cos(m\theta)}{1 - \cos(n\theta)} = \lim_{\theta \to 0} \frac{m^2 \theta^2 / 2}{n^2 \theta^2 / 2} \\\textbf{Simplify the expression} \\\text{The } \frac{1}{2} \text{ terms cancel out, and the } \theta^2 \text{ terms also cancel out:} \\\lim_{\theta \to 0} \frac{m^2}{n^2} = \frac{m^2}{n^2}$$​​