Let $f:\to \mathrm{ℝ}\to (0,\infty )$ be strictly increasing function such that $\lim _{x\to \infty }\frac{f(7x)}{f(x)}=1$. Then, the value of $\lim _{x\to \infty }\left[\frac{f(5x)}{f(x)}-1\right]$ is equal to [JEE Main 2024, 31 Jan (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: c) 0. $f$ is increasing function. $x<5x<7x$ $f(x)<f(5x)<f(7x)$ $\frac{f(x)}{f(x)}<\frac{f(5x)}{f(x)}<\frac{f(7x)}{f(x)}$ $\lim_{x\to\infty}\frac{f(x)}{f(x)}<\lim_{x\to\infty}\frac{f(5x)}{f(x)}<\lim_{x\to\infty}\frac{f(7x)}{f(x)}$ $1<\lim_{x\to\infty}\frac{f(5x)}{f(x)}<1\Rightarrow \lim_{x\to\infty}\frac{f(5x)}{f(x)}=1$ $\lim_{x\to\infty}\left(\frac{f(5x)}{f(x)}-1\right)=0$

Let \(f:\to \mathrm{ℝ}\to (0,\infty )\) be strictly increasing function such that \(\lim _{x\to \infty }\frac{f(7x)}{f(x…

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