Prove $f(x)\sin x$ has a minimum.

Question

We are given $f(x)$, continuous on $[0,\infty)$ such that $\lim_{x\rightarrow\infty} f(x)=\ell $. I already proved the first part, that $g(x)=f(x)\sin x$ is bounded. However I cannot prove that if $\ell=0$ then $g$ has a minimum on $[0,\infty)$.

There's a hint given to look at the case $g\geq0$ first, but I can't see how does that help me, beside knowing that when $\sin x<0$ then $f(x)<0$ (P.S, assuming I proved it, how does it relate to the genral case even?).

Answer 1

1) If $g(x)\ge 0$ then there are some points (when $\sin(x)=0)$ for which $g(x)=0$, and the infimum is attained in those points (e.g. $x=0$).

2) If there is a point $x_0$ for which $g(x_0)=-\epsilon<0$, then from $\lim_{x\rightarrow\infty} g(x)=0$ there is a $M>0$ for which $g(x)>-\epsilon=g(x_0)$ when $x>M$. So the infimum can only be attained in $[0,M]$, and $f$ is continuous on this closed interval, so it attains its minimum.