1. **Stating the problem:** We want to find the limit $\lim_{z \to n\pi} \frac{z - n\pi}{\sin z}$. 2. **Substitution check:** Direct substitution yields $\frac{n\pi - n\pi}{\sin(n\pi)} = \frac{0}{0}$, which is an indeterminate form. 3. **Apply L'Hôpital's rule:** Since the limit yields $\frac{0}{0}$, we differentiate numerator and denominator separately with respect to $z$. - Derivative of numerator: $\frac{d}{dz}(z - n\pi) = 1$ - Derivative of denominator: $\frac{d}{dz}(\sin z) = \cos z$ 4. **Evaluate the limit of the derivatives:** Substitute $z = n\pi$: $$\lim_{z \to n\pi} \frac{1}{\cos z} = \frac{1}{\cos(n\pi)}.$$ 5. **Simplify the cosine term:** We know $\cos(n\pi) = (-1)^n$. 6. **Final answer:** The limit is $$\lim_{z \to n\pi} \frac{z - n\pi}{\sin z} = \frac{1}{(-1)^n} = (-1)^n.$$