Why does $(2^x -1)/x$ not have a vertical asymptote? Why does $(2^x -1)/x$ not have a vertical asymptote as opposed to $(2^x)/x$?
Can you factor in the first expression so that you can cancel out the $x$ term?
 A: This is a really wonderful question!
For small values of $x$, $2^x$ is approximately equal to $1 + (\ln 2)x$.  This is not obvious (unless you know Calculus), but if you graph both functions and zoom in you will see that their graphs are indistinguishable near $x = 0$.
So for small values of $x$,
$$\frac{2^x - 1}{x} \approx \frac {1 + (\ln 2)x - 1}{x} = \ln 2$$
A: mweiss gave already a nice explanation, but let me give a proof requiring as little calculus as I can.
Let $f:\mathbb{R} \setminus \{0\} \to \mathbb R$ be $f(x) = (2^x-1)/x$.
For function $f$ to have vertical asymptote at $x = 0$ it would have to be unbounded near $0$. Our function is not defined at $0$, but there exists limit: $$
\lim_{x\to 0} f(x) = \lim_{x\to 0}\frac{2^x - 2^0}{x - 0} = \left. (2^x)' \right|_{x=0} = \left. (\ln 2\cdot 2^x) \right|_{x=0} =\ln 2.
$$
Let's define $\tilde f:\mathbb R \to \mathbb R$ as $$
\tilde f(x) =\begin{cases}
          f(x) \quad &\text{if} \, x \neq 0 \\
          \ln 2\quad &\text{if} \, x = 0. \\
     \end{cases}
$$
Since $f$ was continuous on its domain, and $\tilde f$ is continuous at $x = 0$, then $\tilde f$ is continuous everywhere. So in particular, it is bounded near $0$ (formally, using Extreme Value Theorem it is bounded on for example $[-1, 1]$). That shows $f$ is bounded near $0$.        $\square$
