Let f : R ( 0 , ? ) be strictly increasing function such that lim x ? f ( 7 x ) f ( x ) = 1 . Then, the value of lim x ? [ f ( 5 x ) f ( x ) - 1 ] is equal to [2024]

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Solution

Q.
Let $f : R \to (0, \infty)$ be strictly increasing function such that $\lim_{x \to \infty} \frac{f (7 x)}{f (x)} = 1 .$ Then, the value of $\lim_{x \to \infty} [\frac{f (5 x)}{f (x)} - 1]$ is equal to [2024]

1 4

2 0

3 7/5

4 1

Ans.
(2)

$For x > 0,$

$x \leq 5 x \leq 7 x \Rightarrow f (x) \leq f (5 x) \leq f (7 x) (∵ f is strictly increasing function)$

$\Rightarrow \frac{f (x)}{f (x)} \leq \frac{f (5 x)}{f (x)} \leq \frac{f (7 x)}{f (x)} \Rightarrow 1 \leq \frac{f (5 x)}{f (x)} \leq \frac{f (7 x)}{f (x)}$

$\Rightarrow 1 \leq \lim_{x \to \infty} \frac{f (5 x)}{f (x)} \leq 1 [∵ \lim_{x \to \infty} \frac{f (7 x)}{f (x)} = 1]$

$∴ \lim_{x \to \infty} \frac{f (5 x)}{f (x)} - 1 = 0$

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