MCQ
$\mathop {{\rm{lim}}}\limits_{x \to 2} \left( {\frac{{\sqrt {1 - {\rm{cos}}\left\{ {2\left( {x - 2} \right)} \right\}} }}{{x - 2}}} \right)=$
  • A
    $\sqrt 2 $
  • B
    -$\;\sqrt 2 $
  • C
    $\frac{1}{{\sqrt 2 }}$
  • does not exist

Answer

Correct option: D.
does not exist
d
$\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {1 - \cos \{ 2(x - 2)\} } }}{{x - 2}}$

$ = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt 2 |\sin (x - 2)|}}{{x - 2}}$

$R.H.L. = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\sqrt 2 \sin (x - 2)}}{{(x - 2)}} =  - \sqrt 2 $

$R.H.L. = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt 2 \sin (x - 2)}}{{(x - 2)}} =  - \sqrt 2 $

Thus $L . H . L . \neq R . H . L$

Hence, $\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {1 - \cos \{ 2(x - 2)\} } }}{{x - 2}}$ does not exist.

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