Using a known limit and the limit laws to evaluate lim (e^(2x)-1)/x

Rewrite the expression by multiplying and dividing by 2: $\frac{e^{2x}-1}{x} = 2\cdot\frac{e^{2x}-1}{2x}.$ This is valid for $x \neq 0$, and it sets up the expression $\frac{e^{2x}-1}{2x}$, which matches the form of the given known limit with the substitution $u = 2x$.

As $x \to 0$, we also have $u = 2x \to 0$, so $\lim_{x\to 0} \frac{e^{2x}-1}{2x} = \lim_{u \to 0} \frac{e^{u}-1}{u} = 1$. Applying the constant multiple rule (a special case of the product rule), $\lim_{x\to 0} \frac{e^{2x}-1}{x} = 2 \cdot \lim_{x\to 0}\frac{e^{2x}-1}{2x} = 2 \cdot 1 = 2.$

The answer is two. The solution rewrote the quotient as two times the quantity e to the two x minus one, over two x, so that the inner piece would match the known limit form exactly. This multiply-and-divide trick let us apply the constant multiple rule afterward.