Evaluating a 0/0 limit using L'Hopital's Rule

Substituting $x=0$ gives numerator $e^0 - e^0 = 0$ and denominator $\sin(0) = 0$, so the limit has the indeterminate form $0/0$. This confirms L'Hopital's Rule applies.

Differentiate numerator and denominator separately: the derivative of $e^{2x}-e^{-2x}$ is $2e^{2x}+2e^{-2x}$, and the derivative of $\sin(4x)$ is $4\cos(4x)$. The new limit is $\lim_{x\to 0} \frac{2e^{2x}+2e^{-2x}}{4\cos(4x)}.$

Substituting $x=0$ now gives $\frac{2(1)+2(1)}{4(1)} = \frac{4}{4} = 1$, which is no longer indeterminate. Therefore the original limit equals $1$.