Limits Question 10
Question 10 - 2024 (31 Jan Shift 2)
Let $\quad \mathrm{f}: \rightarrow \mathrm{R} \rightarrow(0, \infty)$ 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
(1) 4
(2) 0
(3) $7 / 5$
(4) 1
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Answer (2)
Solution
$f: R \rightarrow(0, \infty)$
$\lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$
$\because \mathrm{f}$ is increasing
$\therefore \mathrm{f}(\mathrm{x})<\mathrm{f}(5 \mathrm{x})<\mathrm{f}(7 \mathrm{x})$
$\because \frac{\mathrm{f}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(7 \mathrm{x})}{\mathrm{f}(\mathrm{x})}$
$1<\lim _{x \rightarrow \infty} \frac{f(5 x)}{f(x)}<1$
$\therefore\left[\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}-1\right]$
$\Rightarrow 1-1=0$
