# Question #34fda

Because ${\lim}_{x \to {0}^{-}} f \left(x\right) \ne {\lim}_{x \to {0}^{+}} f \left(x\right)$ and is different from $f \left(0\right) = 0$.

We have that

${\lim}_{x \to {0}^{-}} \frac{\left\mid x \right\mid}{{x}^{2} + 2 x} = {\lim}_{x \to {0}^{-}} \frac{- x}{{x}^{2} + 2 x} = {\lim}_{x \to {0}^{-}} - \frac{1}{x + 2} = - \frac{1}{2}$

and

${\lim}_{x \to {0}^{+}} \frac{\left\mid x \right\mid}{{x}^{2} + 2 x} = {\lim}_{x \to {0}^{+}} \frac{x}{{x}^{2} + 2 x} = {\lim}_{x \to {0}^{+}} \frac{1}{x + 2} = \frac{1}{2}$

The graph of f(x) around zero is