Finding limits of complex functions using Taylor expansion

Question

I am supposed to compute the following limit: $$ \lim_{z \to 0} \frac{(1-\cos z)^2}{(e^z-1-z)\sin^2z} $$

I guess I have to use a Taylor expansion somehow, but I'm not sure what to expand and how, it looks a bit complicated. Any ideas?

Answer 1

$$1-\cos(z)=\frac{z^2}{2}+o(z^2),$$ therefore $$(1-\cos(z))^2=\frac{z^4}{4}+o(z^4).$$ Also, $$\sin(z)=z+o(z),$$ and thus $$\sin^2(z)=z^2+o(z^2).$$

Moreover,

$$e^{z}=1+z+\frac{z^2}{2}+o(z^2),$$ and thus $$\sin^2(z)(e^z-1-z)=\frac{z^4}{2}+o(z^4).$$

I let you conclude.

Answer 2

Writing the first terms that do not cancel out, the expression is

$$\sim\frac{\left(\dfrac{z^2}{2!}\right)^2}{\dfrac{z^2}{2!}z^2}.$$