Is the set of all $f$ such that $\lim_{x\to1^-}f(x) = 0$ an ideal in the ring of functions from $[0,1]\rightarrow \mathbb{R}$?

Question

Is the set of all $f$ such that $\lim_{x\to1^-}f(x) = 0$ an ideal in the ring of functions from $[0,1]\rightarrow \mathbb{R}$? I'm sure about the closure under addition but not quite clear about if $rs\in I \space \forall r\in R,s\in I$ part...

Answer 1

Indeed, let $f(x)=1-x$ and $g(x)=\begin{cases}0,&x\in\Bbb Q\\\frac1{1-x},&x\notin\Bbb Q\end{cases}$. Then $f\in I$ but $g\cdot f\notin I$.

Answer 2

Hint: Use the fact that if $\lim_{x\to 1^{-}} g(x)=a$ and $\lim_{x\to 1^{-}} f(x)=b$, then you have $\lim_{x\to 1^{-}} (g\cdot f)(x)=ab$.

Also note that in the ring of functions from $[0,1]\to \mathbb{R}$, there are problematic functions.