Find all possible values of the following expression: 1/(x^2-2x+3)

$\frac{1}{x^2-2x+3}=\frac{1}{(x-1)^2+2}$

Now we know that $(x-1)^2≥0\,∀\,x∈R$. Thus,

$(x-1)^2+2≥2\,∀\,x∈R$

or $2≤(x-1)^2+2<∞$

or $\frac{1}{2}≥\frac{1}{(x-1)^2+2}>0$

or $\frac{1}{x^2-2x+3}∈\left(0,\frac{1}{2}\right]$