$f(x) = \frac{x \tan x}{1 - \cos x}$ for $x \ne 0$, and the value of $f$ at $x = 0$ is $k$. For which $k$ is $f$ continuous at $x = 0$?

Explanation

1. 01

   $1 - \cos x = 2\sin^{2} \frac{x}{2}$, so $f(x) = \frac{\tan x}{x} \cdot \frac{x^{2}}{2\sin^{2} \frac{x}{2}} = \frac{\tan x}{x} \cdot 2\left(\frac{\frac{x}{2}}{\sin \frac{x}{2}}\right)^{2}$.

2. 02

   As $x \to 0$ this tends to $1 \times 2 \times 1 = 2$, so $k = 2$.