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Let $f:R\rightarrow R$ be a differentiable function having $f(2)=6,f'(2)=\large\frac{1}{48}$.Then $\lim\limits_{x\to 2}\int_6^{f(x)}\large\frac{4t^3dt}{x-2}$ equals


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$\lim\limits_{x\to 2}\int_6^{f(x)}\large\frac{4t^3dt}{x-2}$
It is of the form $\large\frac{0}{0}$
From L Hospital rule,
$\lim\limits_{x\to 2}\large\frac{4(f(x))^3\times f'(x)}{1}$
$\Rightarrow 4(f(2))^3\times f'(2)$
$\Rightarrow 18$
Hence (b) is the correct answer.
answered Dec 24, 2013 by sreemathi.v

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