_ x 1 1 - 2(x - 1) x - 1 - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 - \cos 2(x - 1)} }}{{x - 1}}$

A
Exists and it equals $\sqrt 2 $
B
Exists and it equals $ - \sqrt 2 $
C
Does not exist because $x - 1 \to 0$
✓
Does not exist because left hand limit is not equal to right hand limit

✓

Answer

Correct option: D.

Does not exist because left hand limit is not equal to right hand limit

d
(d) $f(1 + ) = \mathop {\lim }\limits_{h \to 0} f(1 + h) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sqrt {1 - \cos \,\,2h} }}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \,\sqrt 2 \frac{{\sin \,h}}{h} = \sqrt 2 $
$f(1 - ) = \mathop {\lim }\limits_{h \to 0} f(1 - h) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sqrt {1 - \cos \,( - 2h)} }}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \,\sqrt 2 \frac{{\sin \,h}}{{ - h}} = - \sqrt 2 .$
$\therefore $ limit does not exist because left hand limit is not equal to right hand limit.

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