Show that this ring has no identity. 
Let $R=\left \{ g:\Bbb{R}\to \Bbb{R} \mid g \text{ is continuous and } g(1)=0 \right \}$ be a ring. Show that $R$ has no identity.

The answer says there does not exist a function $h(x)\in R$ such that $h(x)=1$, which I don't understand why, since the only condition in $R$ is $g(1)=0$. Please help me understand what I am missing.

The multiplication in $R$ is defined by
$$
fg\colon x\mapsto f(x)g(x)
$$
for $f,g\in R$. Suppose an identity exists, call it $e$.
Suppose $x_0\ne1$; then there is $f\in R$ such that $f(x_0)=1$. Indeed, you can consider $f(x)=(x-1)/(x_0-1)$ as this function is continuous and $f(1)=0$. Since $e$ is the identity, we must have $ef=f$, so in particular
$$
e(x_0)f(x_0)=f(x_0)
$$
that entails $e(x_0)=1$. So the function $e$ must satisfy $e(x)=1$ for every $x\ne1$. As a consequence
$$
\lim_{x\to1}e(x)=1
$$
However, $e\in R$, so $e$ is continuous; this implies $e(1)=1$: a contradiction, because $e\in R$ implies $e(1)=0$.
