(e^(2x) − 1) over x

Computation. For $x\neq0$,
$$\frac{e^{2x}-1}{x}=2\cdot\frac{e^{2x}-1}{2x}.$$
With $u=2x\to0$, $\dfrac{e^u-1}{u}\to1$. The limit is $2$.

Why $\frac{e^u-1}u\to1$? It is exactly the derivative of $\exp$ at $0$: $\frac{e^u-e^0}{u-0}\to\exp'(0)=e^0=1$.