Question

$\underset{\theta \to 0}{\lim }\frac{1-\cos 4\theta }{1-\cos 6\theta }$

Answer 1

$$\begin{gathered} \underset{\mathrm{\theta }\to 0}{\lim }\left[\frac{1-\cos \left(4\mathrm{\theta }\right)}{1-\cos 6\mathrm{\theta }}\right] \\ =\underset{\mathrm{\theta }\to 0}{\lim }\left[\frac{2{\sin }^{2}2\mathrm{\theta }}{2{\sin }^{2}3\mathrm{\theta }}\right]\left\{\because 1-\cos \mathrm{A}=2{\sin }^2\left(\frac{\mathrm{A}}{2}\right)\right\} \\ =\underset{\mathrm{\theta }\to 0}{\lim }\left[\frac{{\sin }^{2}2\mathrm{\theta }}{{\left(2\mathrm{\theta }\right)}^2}\times \frac{{\left(2\theta \right)}^2}{{\displaystyle \frac{{\sin }^{2}3\theta }{{\left(3\theta \right)}^2}}\times {\left(3\mathrm{\theta }\right)}^2}\right] \\ =\underset{\mathrm{\theta }\to 0}{\lim }\left[{\left(\frac{\sin 2\mathrm{\theta }}{2\mathrm{\theta }}\right)}^2\times {\left(\frac{3\mathrm{\theta }}{\sin 3\mathrm{\theta }}\right)}^2\times \frac{4}{9}\right] \\ =\frac{4}{9}\left[\because \underset{X\to 0}{\lim }\frac{\sin x}{x}=1\right] \end{gathered}$$