Question

$\displaystyle\lim_{\theta \rightarrow 0}\frac{4 \theta (\tan \theta -2 \theta \,\tan\,theta)}{(1-\cos\,2\theta)}$ is

• A. $1/\sqrt{2}$
• B. 1/2
• C. 1
• D. 2

Answer 1

Answer: (D)

$ \displaystyle\lim _{\theta \rightarrow 0} \frac{4 \theta(\tan \theta-2 \theta \tan t h e t a)}{1-\cos 2 \theta}$
$= \displaystyle\lim _{\theta \rightarrow 0} \frac{4\left(\theta \tan \theta-2 \theta^{2} \tan \theta\right)}{1-\cos 2 \theta}$ Using L' Hospital's rule
$= \displaystyle\lim _{\theta \rightarrow 0} \frac{4\left(\theta \sec ^{2} \theta+\tan \theta-4 \theta \tan \theta-2 \theta^{2} \sec ^{2} \theta\right)}{2 \sin 2 \theta}$
Again using L' Hospital's rule
$4\left(\sec ^{2} \theta+2 \theta \sec ^{2} \tan \theta+\sec ^{2} \theta-4 \tan \theta\right)$
$= \displaystyle\lim _{\theta \rightarrow 0} \frac{-4 \theta \sec ^{2} \theta-4 \theta \sec ^{2} \theta-4 \theta^{2} \sec ^{2} \theta \tan \theta}{4 \cos 2 \theta}$
$=\frac{4(1+0+1)}{4}=2$