Question

$\underset{\theta \to 0}{\lim }\frac{4\theta (\tan \theta -2\theta \tan theta)}{(1-\cos 2\theta )}$ is

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

Answer 1

Answer: (D)

$\underset{\theta \to 0}{\lim }\frac{4\theta (\tan \theta -2\theta \tan theta)}{1-\cos 2\theta }$
$=\underset{\theta \to 0}{\lim }\frac{4\left(\theta \tan \theta -2{\theta }^2\tan \theta \right)}{1-\cos 2\theta }$ Using L' Hospital's rule
$=\underset{\theta \to 0}{\lim }\frac{4\left(\theta {\sec }^2\theta +\tan \theta -4\theta \tan \theta -2{\theta }^2{\sec }^2\theta \right)}{2\sin 2\theta }$
Again using L' Hospital's rule
$4\left({\sec }^2\theta +2\theta {\sec }^2\tan \theta +{\sec }^2\theta -4\tan \theta \right)$
$=\underset{\theta \to 0}{\lim }\frac{-4\theta {\sec }^2\theta -4\theta {\sec }^2\theta -4{\theta }^2{\sec }^2\theta \tan \theta }{4\cos 2\theta }$
$=\frac{4(1+0+1)}{4}=2$