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