Question

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

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

Answer 1

Answer: (D)

$\underset{x\to 2}{\lim }\frac{\sqrt{1-\cos \left\{2\left(x-2\right)\right\}}}{x-2}$
$=\underset{x\to 2}{\lim }\frac{\sqrt{2}\left|\sin \left(x-2\right)\right|}{x-2}$
$L.H.L.=\underset{x\to 2^{-}}{\lim }\frac{\sqrt{2}\sin \left(x-2\right)}{\left(x-2\right)}=-1$
$R.H.L.=\underset{x\to 2^{+}}{\lim }\frac{\sqrt{2}\sin \left(x-2\right)}{\left(x-2\right)}=1$
Thus $L.H.L.\ne R.H.L.$
Hence, $\underset{x\to 2}{\lim }\frac{\sqrt{1-\cos \left\{2\left(x-2\right)\right\}}}{x-2}$ does not exist.