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