# Evaluating $\lim\limits_{z \to 0} \frac{z\cdot \cos(z)}{\sin(z)}$

I'm reading Boas' chapter on functions of a complex variable, and she's talking here about finding residues. However, I don't understand the evaluation of the limit(below). How is it that we can take $\cos(0)$ out of the limit while $\sin(z)$ and $z$ stay within the limit?

$$\lim_{z \to 0} \frac{z\cdot\cos(z)}{\sin(z)}=\cos (0)\cdot \lim_{z \to 0}\frac{z}{\sin(z)}=1\cdot 1=1$$

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If we do it more slowly, we have $\lim \frac{z\,\cos z}{\sin z} = \lim \cos z \cdot \lim\frac z{\sin z} = \cos(0) \cdot \lim\frac{z}{\sin z}$ ... – martini Jul 18 '12 at 9:57
Is that book Mathematical Methods in the Physical Sciences? – lhf Jul 18 '12 at 10:38
@lhf - yes, it is. – Joebevo Jul 18 '12 at 10:43
The limit of the product is the product of the limits. Proof at PlanetMath. – Raymond Manzoni Jul 18 '12 at 10:48

You can take stuff out of the limits which are smooth (or even continuous) and non-zero about that point. Here $\cos(0) = 1$, but $z(0)$ and $\sin(z)$ although smooth, evaluate to zero.

The reason we can do this is the product rule of limits. We can actually write an intermediary step including $lim_{z\rightarrow0}\cos(z)$ which evaluates to $\cos(0)$. We can't take the other two parts $z(0)$ and $\sin(z)$ as they evaluate to zero and we'd be left with a 0/0.

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I think you CAN take $z$ or $sin(z)$ out of the limit, however, you will not be able to solve the resulting equation. For example, if you take z out of the limit, you get:

$$\lim_{z \to 0} \frac{z\cdot\cos(z)}{\sin(z)}=0 \cdot \lim_{z \to 0}\frac{\cos(z)}{\sin(z)}=0\cdot \infty=???$$

Boas was just smart enough to arrange the equation in a way that will become solvable...

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