Question

$\underset{x\to 0}{\text{lim}}\frac{\cos \left(\sin x\right)-1}{x^2}$ is equal to

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

Answer 1

The correct option is D $-\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{\cos \left(\sin x\right)-1}{x^2}=\underset{x\to 0}{lim}\frac{-2{\sin }^2\left(\frac{\sin x}{2}\right)}{x^2}=-2\underset{x\to 0}{lim}\frac{{\left[\sin \left(\frac{\sin x}{2}\right)\right]}^2.{\left(\frac{\sin x}{2}\right)}^2}{{\left(\frac{\sin x}{2}\right)}^2\times x^2}=-2\cdot (1{)}^2.\frac{1}{4}=-\frac{1}{2}$