Question

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

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

Answer 1

The correct option is B Does not exist

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

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

and substitute $x-2=t$ $\therefore$ as $x\to 2t\to 0$

$\underset{x\to 2}{lim}$ $\left(\frac{\sqrt{2{\sin }^2(x-2)}}{x-2}\right)=\underset{t\to 0}{lim}$ $\frac{\sqrt{2}\left|sint\right|}{t}$

Now in order to remove the modulus we have separately solve the left hand and right hand limit.

$L.H.L$

$\underset{t\to 0^{-}}{lim}$ $\frac{-\sqrt{2}sint}{t}=-\sqrt{2}\underset{t\to 0^{-}}{lim}$ $\frac{sint}{t}=-\sqrt{2}$ (for $t<0$ $\left|sint\right|=-sint$)

$R.H.L$

$\underset{t\to 0^{+}}{lim}$ $\frac{\sqrt{2}sint}{t}=\sqrt{2}\underset{t\to 0^{+}}{lim}$ $\frac{sint}{t}=\sqrt{2}$ ($t>0$, $\left|sint\right|=sint$)

Since, $LHL\ne RHL$ we say that the limit does not exist.

Correct answer : Option B.