Let f: R→ R be a positive increasing function with limx→∞f(3x)f(x)=1,then limx→∞f2xfx=

since f is an increasing function ⇒fx≤f2x≤f3x ⇒1≤f2xfx≤f3xfx ⇒1≤limx→∞f2xfx≤limx→∞f3xfx ⇒ 1≤limx→∞f2xfx≤1 by sandwitch theorem limx→∞f2xfx=1

Let f: R→ R be a positive increasing function with limx→∞f(3x)f(x)=1,then limx→∞f2xfx=

since f is an increasing function ⇒fx≤f2x≤f3x ⇒1≤f2xfx≤f3xfx ⇒1≤limx→∞f2xfx≤limx→∞f3xfx ⇒ 1≤limx→∞f2xfx≤1 by sandwitch theorem limx→∞f2xfx=1

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Let $f : R \to R$ be a positive increasing function with $\lim_{x \to \infty} \frac{f (3 x)}{f (x)} = 1,$ then $\lim_{x \to \infty} \frac{f (2 x)}{f (x)} =$

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$\frac{2}{3}$

$\frac{3}{2}$

answer is D.

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Detailed Solution

$\sin c e f i s a n i n c r e a \sin g f u n c t i o n \Rightarrow f (x) \leq f (2 x) \leq f (3 x) \Rightarrow 1 \leq \frac{f (2 x)}{f (x)} \leq \frac{f (3 x)}{f (x)} \Rightarrow 1 \leq \lim_{x \to \infty} \frac{f (2 x)}{f (x)} \leq \lim_{x \to \infty} \frac{f (3 x)}{f (x)} \Rightarrow 1 \leq \lim_{x \to \infty} \frac{f (2 x)}{f (x)} \leq 1 b y s a n d w i t c h t h e o r e m \lim_{x \to \infty} \frac{f (2 x)}{f (x)} = 1$