Is $f(x)=g(x)(\frac{1}{\sin x }-\frac{1}{ x })$ unifrmly continuous ? where $g:(0, 1)\longrightarrow \mathbb R$ be unifrmly continuous

Question

Let $g:(0, 1)\longrightarrow \mathbb R$ be unifrmly continuous and $f:(0, 1)\longrightarrow \mathbb R$ defined by $f(x)=g(x)(\frac{1}{\sin x }-\frac{1}{ x })$

I was thinking about

1: Is $f(x)=g(x)(\frac{1}{\sin x }-\frac{1}{ x })$ unifrmly continuous ?

2: Is $f(x)=g(x)(\frac{1}{\sin x }-\frac{1}{ x })$ bounded ?

I think $f(x)$ is unifrmly continuous

Answer 1

Hints: $\frac 1 {\sin x} -\frac 1 x \to 0$ as $x \to 0$. Hence this function is bounded and uniformly continuous. [A continuous function on $(0,1)$ is uniformly continuous iff it has finite limits at $0$ and $1$].

Product of two bounded uniformly continuous functions is uniformly continuous (and bounded): Use the fact that $|f(x)g(x)-f(y)g(y)| \leq |f(x)| |g(x)-g(y)| +|g(y)| |f(x)-f(y)|$