One-sided limits of e^(1/x) at x = 0

As $x \to 0^-$, $x$ is a small negative number, so $1/x \to -\infty$. Since $e^u \to 0$ as $u \to -\infty$, we get $\lim_{x \to 0^-} e^{1/x} = 0$. This is a finite limit, so there is no unbounded behavior from the left side.

As $x \to 0^+$, $x$ is a small positive number, so $1/x \to +\infty$. Since $e^u \to +\infty$ as $u \to +\infty$, we get $\lim_{x \to 0^+} e^{1/x} = +\infty$. Because this one-sided limit is infinite, $f$ grows without bound as $x \to 0^+$, which is exactly the condition needed for a vertical asymptote — even though the left-hand limit is finite. Therefore $x=0$ is a vertical asymptote of $f(x)=e^{1/x}$, since at least one one-sided limit there is infinite.