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The value of $\lim\limits_{x\to 0}\large\frac{\sqrt{\Large\frac{1}{2}\normalsize (1-\cos 2x)}}{x}$

$(A)\;1\qquad(B)\;-1\qquad(C)\;0\qquad(D)\;None\;of\;these$

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$\lim\limits_{x\to 0}\large\frac{\sqrt{\Large\frac{1}{2}\normalsize (1-\cos 2x)}}{x}=$$\lim\limits_{x\to 0}\large\frac{\sqrt{(\Large\frac{1}{2}\normalsize .2\sin^2x)}}{x}$
$\qquad\qquad\qquad\;\;= \lim\limits_{x\to 0}\large\frac{|\sin x|}{x}$
LHL=$\lim\limits_{h\to 0}\large\frac{|\sin(0-h)|}{0-h}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{|-\sin h|}{x}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{\sin h}{-h}$$=-1$
RHL=$\lim\limits_{h\to 0}\large\frac{|\sin(0+h)|}{0+h}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{|\sin h|}{h}$
$\Rightarrow 1$
As LHL $\neq$ RHL
The given limit does not exist.
Hence (d) is the correct answer.
answered Jan 4, 2014 by sreemathi.v
 

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