Question

The value of $\underset{\theta \to 0}{lim}\frac{\tan \theta -\sin \theta }{{\theta }^3}$ is

• A. $0$
• B. $2$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{2}$

$\underset{\theta \to 0}{lim}\frac{\tan \theta -\sin \theta }{{\theta }^3}$

$=\underset{\theta \to 0}{lim}\frac{\frac{\sin \theta }{\cos \theta }-\sin \theta }{{\theta }^3}$

$=\underset{\theta \to 0}{lim}\frac{\sin \theta (1-\cos \theta )}{\theta \cdot {\theta }^2\cos \theta }$

$=\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }\frac{2{\sin }^2\theta /2}{{\theta }^2}\cdot \frac{1}{\cos \theta }$ Since $1-\cos \theta =2{\sin }^2\theta /2$

$=\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }2{\left(\frac{\sin \theta /2}{\theta /2}\right)}^2\times {\left(\frac{1}{2}\right)}^2\times \frac{1}{\cos \theta }$

$=1\times 2\times 1^2\times \frac{1}{4}\times \frac{1}{1}$

$=\frac{2}{4}=\frac{1}{2}$.