Question

Evaluate: $\underset{x\to 0}{\lim }\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x}$

• A. $2$
• B. $-1$
• C. $0$
• D. does not exist

Answer 1

The correct option is D

does not exist

Explanation for the correct option:

Find the value of$\underset{x\to 0}{\lim }\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x}$

Consider the given Equation as,

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x} \\ I=\frac{\sqrt{1-\cos 2\left(0\right)}}{\sqrt{2}\left(0\right)} \\ I=\frac{0}{0}\left[\text{Indeterminent Form}\right] \end{gathered}$$

Using the L' hospital rule

$\underset{x\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

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

We know that

$$\begin{gathered} \mathbf{cos}\mathbf{}\mathbf{2}\mathbf{\theta }\mathbf{=}\mathbf{2}\mathbf{}{\mathbf{cos}}^{\mathbf{2}}\mathbf{\theta }\mathbf{-}\mathbf{1} \\ \mathbf{\text{or}} \\ \mathbf{cos}\mathbf{}\mathbf{2}\mathbf{\theta }\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta } \end{gathered}$$

Then,

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{{\displaystyle \sin 2x}}{\sqrt{2}\times \sqrt{1-1+2{\sin }^{2}x}} \\ \Rightarrow I=\underset{x\to 0}{\lim }\frac{{\displaystyle \sin 2x}}{\sqrt{2}\times \sqrt{2}\left|\sin x\right|} \\ \Rightarrow I=\underset{x\to 0}{\lim }\frac{{\displaystyle \sin 2x}}{2\left|\sin x\right|} \end{gathered}$$

$$\begin{gathered} \Rightarrow I=\underset{x\to 0}{\lim }\frac{x\sin 2x}{2x\left|\sin x\right|} \\ \Rightarrow I=\underset{x\to 0}{\lim }\frac{x}{\left|\sin x\right|} \\ RHL \\ I=1and \\ LHL \\ I=-1 \end{gathered}$$

Hence, the correct answer is option D.