Let f:ℝ→ (0, ∞) be strictly increasing function such that _ x → ∞...

Let, $x$ be any poistive real number.

$⇒ x < 5x < 7x$

$⇒ f\left(x\right) < f\left(5x\right) < f\left(7x\right)$

$⇒ \dfrac{f\left(x\right)}{f\left(x\right)} < \dfrac{f\left(5x\right)}{f\left(x\right)} < \dfrac{f\left(7x\right)}{f\left(x\right)}$

$⇒ 1 < \dfrac{f\left(5x\right)}{f\left(x\right)} < \dfrac{f\left(7x\right)}{f\left(x\right)}$

$⇒ \lim_{x \to \infty}1 < \lim_{x \to \infty}\dfrac{f\left(5x\right)}{f\left(x\right)} < \lim_{x \to \infty}\dfrac{f\left(7x\right)}{f\left(x\right)}$

It is given that, $\lim_{x \to \infty}\dfrac{f\left(7x\right)}{f\left(x\right)} = 1$

So, by using Sandwich theorem,

$⇒ \lim_{x \to \infty}\dfrac{f\left(5x\right)}{f\left(x\right)} = 1$

$⇒ \lim_{x \to \infty}\left\{\dfrac{f\left(5x\right)}{f\left(x\right)} - 1\right\} = 0$