Question
$$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 - \cos \,2\left( {x - 1} \right)} }}{{x - 1}}$$
A.
exists and it is $$\sqrt 2 $$
B.
exists and it is $$ - \sqrt 2 $$
C.
does not exist because $$x - 1 \to 0$$
D.
does not exist because $${\text{LH}}\,{\text{lim}} \ne {\text{RH}}\,{\text{lim}}$$
Answer :
does not exist because $${\text{LH}}\,{\text{lim}} \ne {\text{RH}}\,{\text{lim}}$$
Solution :
$$\eqalign{
& {\text{Limit}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left| {\sin \left( {x - 1} \right)} \right|}}{{x - 1}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left| {\sin \,h} \right|}}{h}{\text{ or }}\mathop {\lim }\limits_{h \to 0} \frac{{\left| {\sin \,h} \right|}}{{ - h}} \cr
& \therefore {\text{ RH limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h}}{h} = 1{\text{ and LH limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h}}{{ - h}} = - 1 \cr} $$