# How do you evaluate the limit (1-cosx)/tanx as x approaches 0?

##### 1 Answer
Sep 28, 2016

$0$

#### Explanation:

Using de Moivre's identity

${e}^{i x} = \cos x + i \sin x$

$\frac{1 - \cos x}{\tan} x = \left(1 - \frac{{e}^{i x} + {e}^{- i x}}{2}\right) \frac{{e}^{i x} + {e}^{- i x}}{{e}^{i x} - {e}^{- i x}} =$
$\frac{\left(2 - \left({e}^{i x} + {e}^{- i x}\right)\right)}{2 \left({e}^{i x} - {e}^{- i x}\right)} \left({e}^{i x} + {e}^{- i x}\right) =$
$\frac{{\left({e}^{i \frac{x}{2}} - {e}^{- i \frac{x}{2}}\right)}^{2} \left({e}^{i x} + {e}^{- i x}\right)}{2 \left({e}^{i \frac{x}{2}} + {e}^{- i \frac{x}{2}}\right) \left({e}^{i \frac{x}{2}} - {e}^{- i \frac{x}{2}}\right)} =$
$= \frac{\left({e}^{i \frac{x}{2}} - {e}^{- i \frac{x}{2}}\right) \left({e}^{i x} + {e}^{- i x}\right)}{2 \left({e}^{i \frac{x}{2}} + {e}^{- i \frac{x}{2}}\right)}$

So

${\lim}_{x \to 0} \frac{1 - \cos x}{\tan} x = {\lim}_{x \to 0} \frac{\left({e}^{i \frac{x}{2}} - {e}^{- i \frac{x}{2}}\right) \left({e}^{i x} + {e}^{- i x}\right)}{2 \left({e}^{i \frac{x}{2}} + {e}^{- i \frac{x}{2}}\right)} = 0$