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

Question

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

Answer 1

I assume that you are familiar with limits of sequences.

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.