Jump of an infinite step function Example: Let $0<x$ be real and let $n$ be natural numbers. Consider the sequence $\left\{ x^n  \right\}_{n=1}^\infty$. Whenever $0<x<1$ one finds $x^n \to 0$ as $n\to\infty$. When $x=1$ one finds $x^n\to 1$, and whenever $1<x$ one finds $x^n\to\infty$. So, the sequence $\left\{ x^n  \right\}_{n=1}^\infty$ tends to $0$, $1$ or $\infty$ for positive $x$. We notice here that $x^n$ tends to $1$ at the jump, between values $0$ and $\infty$. This result can be summarized as follows.
$$
\lim\limits_{n\to\infty} x^n
=
\begin{cases}
0\text{, }0<x<1\\
1\text{, }x=1\\
\infty\text{, }1<x
\end{cases}
$$
Question: Let $a_n(x)$ be strictly positive smooth real function. As $n\to\infty$, let $a_n(x)\to 0$ for $0<x<1$, and $a_n(x)\to \infty$ for $1<x$. Define number $L$ by $L=\lim\limits_{n\to\infty} a_n(1)$. Then $0<L<\infty$.
So, in other words, is it true that for all such $a_n(x)$ the magnitude at the jump is strictly positive finite? Or are there sequences $\left\{ a_n(x)  \right\}_{n=1}^\infty$ such that $L=0$ or $L=\infty$?
 A: Instead of having a constant on $[0, 1/n]$, take a line from $(1/n, 1/n)$ to $(-1, 0)$ and then for $x<-1$, keep all $a_n=0$ (do the same thing for the other family of functions). This way the $a_n$ form a family of continuous functions.
Now, fix some integer $n\geq 2$. Since $a_n$ is continuous on $[-2n, 2n]$, there is a polynomial (by Weierstrass approximation) $p_n$ approximating $a_n$ uniformly within $\epsilon_n$ where $\epsilon_n$ is chosen so that $a_n-\epsilon_n>0$ for $x>0$ (exists because $a_n$ is non zero on the positive side).
This way, $p_n>0$ when $x>0$. Next, take a smooth (positive) bump function $\phi_n$ attaining $1$ on $[-n, n]$ with support in $[-2n, 2n]$ and consider $p_n\phi_n$. This is a smooth function and equal to $p_n$ on $[-n, n]$, however it is $0$ for large $|x|$.
It is clear that the sequence $p_n\phi_n$ is smooth, nonnegative on $x\geq 0$ and has the required limits (note that the $\epsilon_n$ converge to $0$). To obtain strictly positive functions, simply consider $p_n\phi_n+1/n$.
This seems to work. Could you elaborate your comment regarding non standard analysis, I am ignorant on the subject.
