Convergence of a sequence of Functions . Let the function sequence  $\{f_n\}$ be defined by $f_n(x)= x - 2 \exp(-nx) $ for $x \in \mathbb{R}$ . Now let $f :\mathbb{R} \rightarrow \mathbb{R} $ be defined by $f(x)= x-2I\{0\}(x)$ for $x \in [0,\infty]$.Then $f$ is  the limiting function of $\{f_n\}$ on the set $[0,\infty)$ and the sequence $\{f_n\}$ does not converge for $x < 0$ . 
My question is : 


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*Why the function $f(x)=x$ for $x > 0 $ is not the limiting function of $\{f_n\}$ , since the term $2\exp(-nx)$ will approach zero as $n \rightarrow \infty$ for $x >0$ and this term will be undefined when $x=0$ since $0 \times -\infty$ undefined ?.


It turns out that if $x>0$, then $f(x)=x$ so this is the limiting function on $(0,+\infty)$. The role of the indicator function of $\{0\}$ is to have a formula with also works for $x=0$.
