Question

Let $f:(1,\infty )\to (0,\infty )$ be a continuous and decreasing function with $\underset{x\to \infty }{lim}\frac{f\left(4x\right)}{f\left(8x\right)}=1$, then $\underset{x\to \infty }{lim}\frac{f\left(6x\right)}{f\left(8x\right)}$ is equal to

• A. $\frac{4}{8}$
• B. $\frac{4}{6}$
• C. $\frac{6}{8}$
• D. $1$

Answer 1

The correct option is D $1$

Asf(x) is continuous and decreasing function on (0,$\infty )$, so f(8x) < f(6x) < f(4x)
So $\underset{x\to \infty }{lim}1<\underset{x\to \infty }{lim}\frac{f\left(6x\right)}{f\left(8x\right)}<\underset{x\to \infty }{lim}\frac{f\left(4x\right)}{f\left(8x\right)}=1$
By squeeze play theorem $\underset{x\to \infty }{lim}\frac{f\left(6x\right)}{f\left(8x\right)}=1$