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

$\begin{array}{1 1}(a)\;equals\;\sqrt 2&(b)\;equals\;-\sqrt 2\\(c)\;equals\;1/\sqrt 2&(d)\;does\;not\;exist\end{array}$

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$\lim\limits_{x\to 2}\large\frac{\sqrt{1-\cos 2(x-2)}}{x-2}=$$\lim\limits_{x\to 2}\large\frac{\sqrt 2\mid\sin(x-2)\mid}{x-2}$
LHL=$\lim\limits_{x\to 2}\large\frac{\sqrt 2\sin(x-2)}{(x-2)}$$=-\sqrt 2$
$RHL=\lim\limits_{x\to 2}\large\frac{\sqrt 2\sin(x-2)}{(x-2)}$$=\sqrt 2$
Thus $LHL\neq RHL$
Hence $\lim\limits_{x\to 2}\large\frac{\sqrt{1-\cos 2(x-2)}}{x-2}$ does not exist.
Hence (d) is the correct answer.
answered Dec 23, 2013 by sreemathi.v
 

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