Let a function $f:(0,\infty )\to [0,\infty )$ be defined by $f\left(x\right)=|1-\frac{1}{x}|$. Then, $f$ is 

We have,   $f\left(x\right)=\frac{|x-1|}{x}=\{\begin{array}{l}-\frac{(x-1)}{x},if0<x\le 1\\ \frac{x-1}{x},ifx>1\end{array}$

$=\{\begin{array}{l}\frac{1}{x}-1,if0<x\le 1\\ 1-\frac{1}{x},ifx>1\end{array}$

Now, let us draw the graph of   $y=f\left(x\right)$
 Note that when $x\to 0,$ then $f\left(x\right)\to \infty$, when $x=1$, then $f\left(x\right)=0$, and when $x\to \infty ,$ then  $f\left(x\right)\to 1$

Clearly,  $f\left(x\right)$ is not injective because if $f\left(x\right)<1,$ then f is many one, as shown in figure.
          Also, $f\left(x\right)$ is not surjective because range of $f\left(x\right)$ is $[0,\infty ]$  and but in problem co-domain is $(0,\infty )$, which is wrong.  

$\therefore f\left(x\right)$ is neither injective nor surjective