Show that sawtooth function $f$ is continous, where $f$ is given by

$f(n) = \left\{ 
 \begin{array}{l l}
 x-2n & \quad \text{if } {2n \leq{x} \leq{2n+1}}, {n \in \mathbb{Z}}\\
 2n-x & \quad \text{if } {2n-1 \leq{x} \leq{2n}}, {n \in \mathbb{Z}}
 \end{array} \right.$ 

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I know saw tooth is nowhere differentiable

but nothing i can do except taking the limit around $2n^{+}$ and $2n^{-}$

$\lim\_{x \to 2n^{-}} (x-2n) = 0 = \lim\_{x \to 2n^{+}} (2n-x)$

which is super dumb