Question

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

• A. $\frac{4}{9}$
• B. $\frac{9}{4}$
• C. $\frac{9}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{4}{9}$

$\underset{\theta \to 0}{lim}\frac{1-\cos 4\theta }{1-\cos 6\theta }$
$=\underset{\theta \to 0}{lim}\frac{1-\cos 4\theta }{1-\cos 6\theta }\times \frac{1+\cos 4\theta }{1+\cos 4\theta }\times \frac{1+\cos 6\theta }{1+\cos 6\theta }$
$=\underset{\theta \to 0}{lim}\frac{{\sin }^{2}4\theta }{{\sin }^{2}6\theta }$
$=\frac{4^2}{6^2}\times {\left(\underset{\theta \to 0}{lim}\frac{\sin 4\theta }{4\theta }\right)}^2\times {\left(\underset{\theta \to 0}{lim}\frac{6\theta }{\sin 6\theta }\right)}^2$
$=\frac{4}{9}$