Question

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

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

Answer 1

The correct option is B

$\frac{1}{2}$

Explanation of the correct option.

Compute the required value.

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

$$\begin{gathered} =\frac{0+1-1}{0} \\ =\frac{0}{0} \end{gathered}$$

Since it is $\frac{0}{0}$ form, apply L.Hospital's law,

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

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

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