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Q. $\displaystyle\lim_{\theta \to 0} \frac{\sin\, m^2 \theta}{\theta}$ is equal to :

Limits and Derivatives

Solution:

Consider $\displaystyle\lim_{\theta \to 0} \frac{\sin\, m^2 \theta}{\theta}$
= $\displaystyle\lim_{\theta \to 0} \left(\frac{\sin\, m^2 \theta}{m^2\theta}\right) m^2 = 1 \times m^2 = m^2$