Show that sawtooth function $f$ is continous, where $f$ is given by
$f(n) = \left\{
\begin{array}{l l}
x-2n & \quad \text{if } {2n \leq{x} \leq{2n+1}}, {n \in \mathbb{Z}}\\
2n-x & \quad \text{if } {2n-1 \leq{x} \leq{2n}}, {n \in \mathbb{Z}}
\end{array} \right.$
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
Best Answer
The function $$f(x) = \left\{ \begin{array}{l l} x-2n, & \quad \text{if } {2n \leq{x} \leq{2n+1}}, \;\; {n \in \mathbb{Z}}\\ 2n-x, & \quad \text{if } {2n-1 \leq{x} \leq{2n}}, \;\; {n \in \mathbb{Z}} \end{array} \right.$$ is not differentiable only at points $x_n=n, \;\; {n \in \mathbb{Z}},$ but everywhere continuous. You are right that $\lim\limits_{x \to 2n^{-}} f(x) = 0 = \lim\limits_{x \to 2n^{+}} f(x).$ In addition,$\lim\limits_{x \to (2n+1)^{-}} f(x) = 1 = \lim\limits_{x \to (2n+1)^{+}} f(x)$ and $f(x)$ is continuous on every open interval $(2n,\; 2n+1).$