# Pointwise convergence of $h_{n}(x)$ on [0,$\infty$)

I know that it converges pointwise to $$1$$ if $$x>0$$ and to $$0$$ if $$x=0$$ using limits . But I am struggling to show this formally. Any help would be greatly appreciated . Thanks

If $$x>0,$$ then by the fact that the limit of $$\{\frac{1}{n}\}$$ is $$0,$$ for $$\epsilon =x$$, the definition of limit implies that there exists $$n_0=n_0(x)\in \mathbb{N}$$ such that $$\forall n\geq n_0$$ we have $$x>\frac{1}{n}.$$ Then, by definition $$h_n(x)=1,\ \forall n\geq n_0.$$ Consequently, $$h_n(x)\rightarrow 1$$ in this case. Recall, that for $$x$$ fixed this is just the limit of a sequence, whose terms are always equal to $$1$$ after some point. For $$x=0$$ it is $$h_n(0)=0,\forall n\in \mathbb{N}$$ by the definition. So, $$h_n(0)\rightarrow 0$$ as $$n$$ approaches infinity.