Distribution function? Let $F(x) = e^{-1/x}$ for $x>0$ and $F(x)=0$ for $x\leq0$. Now I am investigating if $F$ is a distribution function. Say:
\begin{align}
\int\limits_0^\infty e^{-1/x} \, dx = \left[ \vphantom{\frac11} -x \cdot e^{-1/x} \right]^{x=\infty}_{x=0}.
\end{align}
But how to evaluate those integrands? Such that the integral will be equal to $1$, or not.
 A: $x$ is positive. As $x$ gets bigger, $1/x$ gets smaller.  As $1/x$ gets smaller, $-1/x$ gets bigger.  As $-1/x$ gets bigger, $e^{-1/x}$ gets bigger.
This $F$ is an increasing function on $(0,\infty)$.  Moreover, since $e^{-1/x}$ is always positive when $x$ is real, the values of $F$ on $(0,\infty)$ are bigger than their values on $(-\infty,0]$.
So $F$ is an everywhere non-negative non-decreasing function.  It is clear that it is continuous everywhere, except that we need to examine what it does at $0$ and see whether it satisfies the requirement of being continuous from the right and having a limit from the left.  Having a limit from the left is trivial in this case.  To be continuous from the right, it will be necessary that $\lim\limits_{x\,\downarrow\,0}F(x)=0$, since $F(0)=0$.
As $x$ decreases to $0$, $1/x$ increases to $\infty$, so $-1/x$ decreases to $-\infty$, so $e^{-1/x}$ decreases to $0$.  Thus $F$ is actually continuous everywhere.
It is clear that $F(x)\to0$ as $x\to-\infty$.  There remains only the question of whether $F(x)\to1$ as $x\to+\infty$. As $x$ grows, $1/x$ approaches $0$, so $-1/x$ approaches $0$, so $e^{-1/x}$ approaches $e^0=1$.
