If $f$ is increasing toward $1$, then $\sup\{f(x)\sin x \}=1$

Question

Suppose $f$ is an increasing monotone function in $(0,\infty)$.

If

$$\lim_{x \to \infty} f(x)=1$$ then $$\sup\{f(x)\sin x\mid x>0\}=1$$

I am not really sure how to approach this, any help will be appreciated.

Answer 1

$\sin (x)=1$ iff $x=\dfrac{1}{2}(\pi + 4n \pi)$, $n \in \mathbb{N}$, so define $\{ x_n\}_{n=1}^\infty \subset \mathbb{R}$ by $x_n = \dfrac{1}{2}(\pi + 4n \pi)$, then $$\sup g(x_n) = \sup f(x_n)\sin(x_n)=\sup f(x_n)\cdot 1 = \sup f(x_n) $$ and then conclude!