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

Answer: (A)

$\underset{\theta \to 0}{\lim }\frac{1-\cos 4\theta }{1-\cos 6\theta }$
$=\underset{\theta \to 0}{\lim }\frac{2{\sin }^{2}2\theta }{2{\sin }^{2}3\theta }$
$=\underset{\theta \to 0}{\lim }\frac{{\sin }^{2}2\theta }{4{\theta }^2}\cdot \frac{1}{\frac{{\sin }^{2}3\theta }{9{\theta }^2}}\cdot \frac{4{\theta }^2}{9{\theta }^2}$
$=\underset{\theta \to 0}{\lim }{\left(\frac{\sin 2\theta }{2\theta }\right)}^2\cdot \frac{1}{{\left(\frac{\sin 3\theta }{3\theta }\right)}^2}\cdot \frac{4}{9}$
$=\frac{4}{9}\left[\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right]$