Mostre que a função, apesar de contínua, não tem derivada em $x = 0$

\[\begin{array}{*{20}{c}}
{f\left( x \right)}& = &{\left\{ {\begin{array}{*{20}{c}}
x& \Leftarrow &{x > 0} \\
{ – {x^2}}& \Leftarrow &{x \leqslant 0}
\end{array}} \right.}
\end{array}\]

Calculemos as derivadas laterais no ponto $x = 0$:

\[\begin{array}{*{20}{l}}
{\begin{array}{*{20}{l}}
{f’\left( {{0^ – }} \right)}& = &{\mathop {\lim }\limits_{h \to {0^ – }} \frac{{f\left( {0 + h} \right) – f\left( 0 \right)}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to {0^ – }} \frac{{ – {{\left( {0 + h} \right)}^2} – 0}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to {0^ – }} \frac{{ – {h^2}}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to {0^ – }} \left( { – h} \right)} \\
{}& = &0
\end{array}}&{}&{}&{}&{\begin{array}{*{20}{l}}
{f’\left( {{0^ + }} \right)}& = &{\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {0 + h} \right) – f\left( 0 \right)}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left( {0 + h} \right) – 0}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to {0^ + }} \frac{h}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to {0^ + }} \left( 1 \right)} \\
{}& = &1
\end{array}}
\end{array}\]

Como $f’\left( {{0^ – }} \right) \ne f’\left( {{0^ + }} \right)$, então a função $f$ não tem derivada no ponto de abcissa $0$, apesar de contínua.