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Let $f(x)=4$ and $f'(x)=4$. Then $\lim\limits_{x\to 2}\large\frac{xf(2)-2f(x)}{x-2}$ is given by

$(a)\;2\qquad(b)\;-2\qquad(c)\;-4\qquad(d)\;3$

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Apply LH Rule
We have $\lim\limits_{x\to 2}\large\frac{xf(2)-2f(x)}{x-2}$
$\Rightarrow f(2)-2f'(x)=f(2)-2f'(2)$
$\Rightarrow 4-2\times 4$
$\Rightarrow 4-8$
$\Rightarrow -4$
Hence (c) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 
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