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Let $f : R\rightarrow R$ be a positive increasing function with $\lim\limits_{x\to 0}\large\frac{f(3x)}{f(x)}$$=1$. Then $\lim\limits_{x\to \infty}\large\frac{f(2x)}{f(x)}$ = ?

$(a)\;\large\frac{2}{3}$$\qquad(b)\;\large\frac{3}{2}$$\qquad(c)\;3\qquad(d)\;1$

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$f(x)$ is a positive increasing function $0 < f(x) < f(2x) < f(3x)$
$\Rightarrow 0 < 1< \large\frac{f(2x)}{f(x)}$$ < \large\frac{f(3x)}{f(x)}$
$\Rightarrow \lim\limits_{x\to \infty}1 \leq \lim\limits_{x\to \infty}\large\frac{f(2x)}{f(x)}$$\leq \lim\limits_{x\to \infty}\large\frac{f(3x)}{f(x)}$
By sandwich theorem
$\lim\limits_{x\to 0}\large\frac{f(2x)}{f(x)}$$=1$
Hence (d) is the correct answer.
answered Dec 23, 2013 by sreemathi.v
 

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