Question
What is $$\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {1 - \cos \,x} }}$$ equal to ?
A.
$$\sqrt 2 $$
B.
$$ - \sqrt 2 $$
C.
$$\frac{1}{{\sqrt 2 }}$$
D.
Limit does not exist
Answer :
Limit does not exist
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {1 - \cos \,x} }} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {1 - \left( {1 - 2\,{{\sin }^2}\frac{x}{2}} \right)} }} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {2\,{{\sin }^2}\frac{x}{2}} }} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{x}{{\left| {\sin \frac{x}{2}} \right|}} \cr
& {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( {0 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{x}{{\left| {\sin \frac{x}{2}} \right|}} \cr
& = - \frac{1}{{\sqrt 2 }}\mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\frac{h}{2}} \right)}}{{\sin \frac{h}{2}}} \cr
& = \frac{1}{{\sqrt 2 }} \times 2 \times 1\,\,\,\left( {\because \,\mathop {\lim }\limits_{\theta \to 0} \frac{\theta }{{\sin \,\theta }} = 1} \right) \cr
& = \sqrt 2 \cr
& {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( {0 + 0} \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) \cr
& = \frac{1}{{\sqrt 2 }}\mathop {\lim }\limits_{h \to 0} \frac{{2\left( {\frac{h}{2}} \right)}}{{\sin \frac{h}{2}}} \cr
& = \frac{1}{{\sqrt 2 }} \times 2 \times 1 \cr
& = \sqrt 2 \cr
& {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = \sqrt 2 \cr
& {\text{Therefore limit does not exist}}{\text{.}} \cr} $$