Can you detail $1-\delta <1-xC<1+\delta C\Rightarrow |1-xC|<1+\delta C$?
Hint: if $f$ were continuous at $x=0$, then for $\varepsilon=1$ we would find a $\delta>0$ such that if $|x-0|<\delta$ then $|f(x)-f(0)|\leq 1$. Notice that for $n$ large enough, $\frac 1n<\delta$.