Question

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

​(a) $\frac{4}{9}$

(b) $\frac{1}{2}$

(c) $-\frac{1}{2}$

(d) –1

Answer 1

$$\begin{gathered} \begin{array}{c}\begin{array}{c}\lim \\ \theta \to 0\end{array}\end{array}\frac{1-\cos 4\theta }{1-\cos 6\theta } \\ \left(\mathrm{since}1-\cos 2x=2{\sin }^{2}x\right) \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ \theta \to 0\end{array}\frac{2{\sin }^{2}2\theta }{2{\sin }^{2}3\theta } \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ \theta \to 0\end{array}\frac{{\sin }^{2}2\mathrm{\theta }}{{\sin }^{2}3\mathrm{\theta }} \\ =\begin{array}{c}\lim \\ \theta \to 0\end{array}\frac{{\sin }^{2}2\theta }{{\left(2\theta \right)}^2}\times \frac{{\left(3\theta \right)}^2}{{\sin }^{2}3\theta }\times \frac{{\left(2\theta \right)}^2}{{\left(3\theta \right)}^2} \\ =\begin{array}{c}\lim \\ \theta \to 0\end{array}\frac{{\sin }^{2}2\theta }{{\left(2\theta \right)}^2}\times \frac{{\left(3\theta \right)}^2}{{\sin }^{2}3\theta }\times \frac{4}{9} \\ =\frac{4}{9}\left(\begin{array}{c}\lim \\ \theta \to 0\end{array}\frac{{\sin }^{2}2\theta }{{\left(2\theta \right)}^2}\right)\left(\begin{array}{c}\lim \\ \theta \to 0\end{array}\frac{{\left(3\theta \right)}^2}{{\sin }^{2}3\theta }\right) \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ \theta \to 0\end{array}\frac{1-\cos 4\theta }{1-\cos 6\theta }=\frac{4}{9}\times 1\times 1=\frac{4}{9} \end{gathered}$$

Hence, the correct answer is option A.