Showing that $\lim_{z\to0}z\,\frac{\cos(z/2)}{\sin(z/2)}=2$

Question

Here is an integration using residue thm that I want to understand the final answer of it:

I do not understand why $$\lim_{z\to0}z\,\frac{\cos(z/2)}{\sin(z/2)}=2$$ Could someone explain this to me please?

Answer 1

Start with the well-known fact that $$\lim_{z\to 0} \frac{\sin z}{z} = 1$$ Then $$\lim_{z\to 0} \frac{z}{\sin z} = 1$$ as well. So $$\lim_{z\to 0} \frac{z/2}{\sin z/2} = 1$$ and therefore $$\lim_{z\to 0} \frac{z}{\sin z/2} = 2$$

Finally, combine this with the fact that $\lim_{z\to 0} \cos(z/2) = \cos 0 = 1$ to get $$\lim_{z\to 0} \frac{z \cos(z/2)}{\sin (z/2)} = 2$$

Answer 2

$$\lim_{z \to 0} \frac{z\cos(z/2)}{\sin(z/2)}=\lim_{z \to 0}\cos(z/2) \times \lim_{z \to 0}\frac{z}{\sin(z/2)}\stackrel{L'H}{=}1\times\lim_{z \to 0}\frac{1}{\frac{1}{2}\cos(z/2)}=2 $$

Answer 3

When $x$ is close to zero, $\cos x$ is approximately $1$, and $\sin x$ is approximately $x$.

So for small $z$, $$z\frac{\cos (z/2)}{\sin(z/2)}$$ is approximately $$z\frac1{z/2}=2.$$

Alternatively, use L'Hospital's rule.