Q122 / 200 · Manipal (MET) 2010 · Mathematics
$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{\sqrt{2} \cdot x}$ is
- A1
- B-1
- Czero
- Correct answerDdoes not exist
Solution
$\begin{gathered}\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{\sqrt{2} \cdot x}=\lim _{x \rightarrow 0} \frac{\sqrt{1-\left(1-2 \sin ^2 x\right)}}{\sqrt{2} \cdot x} \\ \quad=\lim _{x \rightarrow 0} \frac{\sqrt{2 \sin ^2 x}}{\sqrt{2} \cdot x}=\lim _{x \rightarrow 0} \frac{|\sin x|}{x}\end{gathered}$
Let $f(x)=\frac{|\sin x|}{x}$
Then,
$f(0+0)=\lim _{h \rightarrow 0} \frac{|\sin (0+h)|}{h}=\lim _{h \rightarrow 0} \frac{\sin h}{h}=1$
and
$f(0-0)=\lim _{h \rightarrow 0} \frac{|\sin (0-h)|}{-h}=\lim _{h \rightarrow 0} \frac{\sin h}{-h}=-1$
$\because f(0+0) \neq f(0-0)$
$\therefore$ The limit does not exist.
Hence, the correct answer is option: 'D'.