Limit of $\frac{(e^{\sin(z)} - 1)^3}{\sin(z) - z}$ as $z \to 0$

Question

I thought about solving this woth L'Hopital's rule, but we never covered the rule for complex limits in class. I can't find a hidden derivative and writing everything out as polynomial also doesn't even feel like it's worth the time.

Answer 1

$$L=\lim_{z \to 0}\frac{(e^{\sin(z)} - 1)^3}{\sin(z) - z}=\lim_{z \to 0} \left(\frac{e^{\sin z}-1}{\sin z}\right)^3\frac{\sin^3z}{\sin z-z}=\lim_{t \to 0} \left(\frac{e^{t}-1}{t}\right)^3 \lim_{z\to 0} \frac{z^3+O(z^4)}{-z^3/6+ O(z^5)}$$ $$\implies L=-6$$