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$\lim\limits_{x\to 2}\large\frac{2^x+2^{3-x}-6}{\sqrt{2^{-x}-2^{1-x}}}$ is

$(a)\;2\qquad(b)\;4\qquad(c)\;8\qquad(d)\;16$

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$\lim\limits_{x\to 2}\large\frac{2^x+2^{3-x}-6}{\sqrt{2^{-x}-2^{1-x}}}=$$\lim\limits_{x\to 2}\large\frac{(2^x)^2-6.2^x+2^3}{\sqrt{2^x-2}}$
Multiplying numerator & denominator by $2^x$
$\Rightarrow \lim\limits_{x\to 2}\large\frac{(2^x-4)(2^x-2)(\sqrt{2^x}+2)}{(\sqrt{2^x}-2)(\sqrt{2^x}+2)}$
$\Rightarrow \lim\limits_{x\to 2}\large\frac{(2^x-4)(2^x-2)(\sqrt{2^x}+2)}{2^x-4}$
$\Rightarrow \lim\limits_{x\to 2}(2^x-2)(\sqrt{2^x}+2)$
$\Rightarrow (2^2-2)(\sqrt{2^2}+2)$
$\Rightarrow 8$
Hence (c) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 
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