$f(2x) = \left\{ \begin{array}{l}1\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,x < 0\\\frac{{1 - [2x]}}{{1 + [2x]}},\,\,x > 0\end{array} \right.$==> $f(2x) = \left\{ \begin{array}{l}1\,\,\,\,\,\,\,\,\,\,,\,\,x < 0\\1\,\,\,\,\,\,\,\,\,\,,\,\,0 \le x < \frac{1}{2}\\0\,\,\,\,\,\,\,\,\,\,,\,\,\frac{1}{2} \le x \le 1\\ - \frac{1}{3}\,\,\,\,\,,\,\,\,1 \le x < \frac{3}{2}\end{array} \right.$
==> $f(x)$, for all values of $x$ where $x < \frac{1}{2}$ a continous function and for $x = \frac{1}{2}$ and $x = 1$ $f(x)$ be a discontinous function.
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