Let f : rightarrow R rightarrow(0, ∞) be strictly increasing function such that lim x rightarrow ∞ f(7 x)/f(x)=1 . Then, the value of lim x rightarrow ∞ [f(5 x)/f(x)-1 ] is equal to :

Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 31 January, Shift 2 — Question 10

Let $\mathrm{f}: \rightarrow \mathrm{R} \rightarrow(0, \infty)$ be strictly increasing function such that $\lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$ . Then, the value of $\lim _{x \rightarrow \infty}\left[\frac{f(5 x)}{f(x)}-1\right]$ is equal to :

Option A:
4
Option B:
0
Correct
Option C:
$7 / 5$
Option D:
1

Answer: B

Step-by-step solution

$\mathrm{f}: \mathrm{R} \rightarrow(0, \infty)$

$\lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$

is increasing $\therefore \mathrm{f}(\mathrm{x})<\mathrm{f}(5 \mathrm{x})<\mathrm{f}(7 \mathrm{x})$

$\because \frac{f(x)}{f(x)}<\frac{f(5 x)}{f(x)}<\frac{f(7 x)}{f(x)}$

$1<\lim _{x \rightarrow \infty} \frac{f(5 x)}{f(x)}<1$

$\therefore\left[\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}-1\right]$

$\Rightarrow 1-1=0$ .

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Sandwich or Squeeze theorem

Answer: B

Step-by-step solution

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