Question

The value of $\displaystyle \lim_{\theta\to0} \frac{1 - \cos 4\theta}{1-\cos 6\theta}$ is

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

Answer 1

Answer: (A)

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