Question

Evaluate: $\underset{x\to 2}{\lim }\left(\frac{\left(\sqrt{1-\cos 2\left(x-2\right)}\right)}{\left(x-2\right)}\right)$

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. does not exist

Answer 1

The correct option is D

does not exist

Explanation for correct option:

Find the value of $\underset{x\to 2}{\lim }\left(\frac{\left(\sqrt{1-\cos 2\left(x-2\right)}\right)}{\left(x-2\right)}\right)$

Consider the given Equation as

$I=\underset{x\to 2}{\lim }\left(\frac{\left(\sqrt{1-\cos 2\left(x-2\right)}\right)}{\left(x-2\right)}\right)$

We know that

$\cos 2\theta =1-2{\sin }^2\theta$

Then, on substituting we have,

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

We know that

$\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$

Then, Right hand limt is

$I=\sqrt{2}$

Similarly, left limitis is,

$$\begin{gathered} I=\underset{x\to 2^{\_}}{\lim }-\sqrt{2}\frac{\sin \left(x-2\right)}{\left(x-2\right)} \\ I=-\sqrt{2} \end{gathered}$$

Since, $\text{LHS}\ne \text{RHS}$, the limit doesn't exist.

Hence, the correct answer is Option D