Can I get a sequence of bounded functions converging pointwise to $f(x)=1/x$ for $x$ non zero and $0$ for $x$ zero?

Question

How do I construct an explicit sequence of bounded functions converging pointwise to $f(x)=1/x$ for non zero $x$ and $f(0)=0$. It would be better if someone may find a continuous and even better if someone gives a differentiable sequence of functions on $\Bbb R$ converging to $f$ pointwise.

Answer 1

The functions $$ f_n(x) = \frac{nx}{1+nx^2} $$ are (infinitely often) differentiable on $\Bbb R$. Each $f_n$ is bounded: $$ |f_n(x)| = \frac{\sqrt{nx^2}}{1+nx^2} \sqrt n \le \frac 12 \sqrt n \, , $$ and $f_n(x) \to f(x)$ pointwise for all $x$.

Graph of $f_5, f_{10}, f_{40}$ together with the graph of $1/x$:

Answer 2

How about: $$ f_n(x) = \begin{cases} 1/x, & \mbox{if } x>\frac{1}{n} \\ n^2\cdot x, & \mbox{if } 0\leq x\leq \frac{1}{n}\end{cases} $$ This is continous, though not differentiable, but you can get the idea on how to do it(What I did is just join the point $(\frac{1}{n},n)$ with $(0,0)$ in a continous manner, sure you can come about with a differentiable way and even $\mathcal{C}^{\infty}$)