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

$\begin{array}{1 1}(a)\;\text{exists and it equals } \sqrt 2 \\(b)\;2\text{exists and it equals } -\sqrt 2\\(c)\;\text{does not exists because }x-1\to 0\\(d)\;\text{does not exists because } LHL\neq RHL\end{array}$

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Putting $x=1+h$
So that when $x\to 1,h\to 0$
Given limit=$\lim\limits_{h\to 0}\large\frac{\sqrt{1-\cos 2h}}{h}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{\sqrt 2\sqrt{\sin^2h}}{h}$
$\Rightarrow \sqrt{2}\lim\limits_{h\to 0}\large\frac{|\sin h|}{h}$ which does not exist.
Hence (c) is the correct answer.
answered Dec 30, 2013 by sreemathi.v

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