Question

The value of $\underset{x\to 0}{\lim }\frac{\left[1-\cos \left(2x\right)\right]}{x^2}$ is

• A. dose not exist.
• B. infinite
• C. $0$
• D. $2$

Answer 1

The correct option is D

$2$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to 0}{\lim }\frac{\left[1-\cos \left(2x\right)\right]}{x^2}$

$$\begin{gathered} =\underset{x\to 0}{\lim }\frac{2{\sin }^2\left(x\right)}{x^2} \\ =\underset{x\to 0}{\lim }2{\left(\frac{\sin \left(x\right)}{x}\right)}^2 \\ =2\times 1 \\ =2 \end{gathered}$$

Therefore, $\underset{x\to 0}{\lim }\frac{\left[1-\cos \left(2x\right)\right]}{x^2}=2$.

Hence option $\left(D\right)$ is the correct option.