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The value of $f(0)$ so that $f(x)=\large\frac{(4^x-1)^3}{\sin\big(\Large\frac{x}{4}\big) \normalsize {\log}(1+\Large\frac{x^2}{3}\big)}$ is continuous everywhere is

$\begin{array}{1 1}(a)\;3(\log 4)^3&(b)\;4(\log 4)^3\\(c)\;12(\log 4)^3&(d)\;15(\log 4)^3\end{array}$

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$\lim\limits_{x\to 0}f(x)=\lim\limits_{x\to 0}\large\frac{(4^x-1)^3}{\sin(\Large\frac{x}{4})\log\big(1+\Large\frac{x^3}{3}\big)}$
$\Rightarrow \lim\limits_{x\to 0}\big(\large\frac{4^x-1}{x}\big)^3\frac{4.x/4}{\sin x/4(1/x^2)\log(1+\Large\frac{x^2}{3})}$
$\Rightarrow \lim\limits_{x\to 0}\big(\large\frac{4^x-1}{x}\big)^3$$4\lim\limits_{x\to 0}\large\frac{x/4}{\sin x/4}\frac{1}{\lim\limits_{x\to 0}\Large\frac{1}{3}\large \log (1+\Large\frac{x^2}{3})^{3/x^2}}$
$\Rightarrow (\log 4)^3.4.1.\large\frac{1}{\Large\frac{1}{3}.\large1}$
$\Rightarrow 12(\log 4)^3$
Hence (c) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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