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JEE Main 2010
Exam
JEE Main
Year
2010
Subject
Mathematics
Chapter
Limits

JEE Main 2010 · Mathematics · Limits

Let f: R arrow R be a positive increasing function with _ x arrow ∞...

Let f:R→Rf: R \rightarrow R be a positive increasing function with lim⁡x→∞f(3x)f(x)=1\lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}=1. Then lim⁡x→∞f(2x)f(x)=\lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=

Options

  1. A23\frac{2}{3}
  2. B32\frac{3}{2}
  3. C3
  4. D1

Correct answer

D. 1

Step-by-step solution

f(x)f(x) is a positive increasing function ⇒0<f(x)<f(2x)<f(3x)⇒0<1<f(2x)f(x)<f(3x)f(x)⇒lim⁡x→∞1≤lim⁡x→∞f(2x)f(x)≤lim⁡x→∞f(3x)f(x)\begin{aligned}& \Rightarrow 0 < f(x) < f(2 x) < f(3 x) \\ & \Rightarrow 0 < 1 < \frac{f(2 x)}{f(x)}< \frac{f(3 x)}{f(x)}\\ & \Rightarrow \lim _{x \rightarrow \infty}1 \leq \lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)} \leq \lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)} \end{aligned}By sandwich theorem. \Rightarrow lim⁡x→∞f(2x)f(x)=\lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=1
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