(2^x − 1) over x

Idea. Every exponential $a^x$ is secretly $e^{x\ln a}$, and we know how to handle $e^u$ near $0$.

Computation. Since $2^x=e^{x\ln2}$,
$$\frac{2^x-1}{x}=\ln2\cdot\frac{e^{x\ln2}-1}{x\ln2}.$$
With $u=x\ln2\to0$, the fraction tends to $1$. The limit is $\ln2$.

Remark. More generally $\dfrac{a^x-1}{x}\to\ln a$ for every $a>0$. It also says that the derivative of $a^x$ at $0$ is $\ln a$, which is how the natural logarithm shows up in the derivative of $a^x$.