La función
$$f(x)= \left\{ \begin{array}{lcc} \frac{x_1x_2^2}{x_1^2+x_2^4} & si & \vec{x} \neq \vec{0} \\ \\ 0 & si & \vec{x} = \vec{0} \end{array} \right.$$
no es continua en $\vec{x}=\vec{0}$

ya que para cada $\varepsilon > 0$ debe existir un $\delta > 0$ tal que $|\vec{x}|<\delta$ $\Rightarrow$ $|f(\vec{x})-f(\vec{0})|<\varepsilon$
para cualquier $\vec{x}$ que pertenezca al entorno esfèrico (en este caso circular) definido por $\delta$.

Pero para $\vec{x}=(h^2,h)$ co $h\neq0$ se tiene

$$|f(\vec{x})-f(\vec{0})|=\frac{h^4}{h^4+h^4}=\frac{1}{2}$$
con lo que para $\varepsilon\leq\frac{1}{2}$ no existe ningún $\delta$ que cumpla la condicion.

Sin embargo,la función tiene en dicho punto derivadas en cualquier dirección:

Sea $\vec{u}=u_1\vec{e_1}+u_2\vec{e_2}$

$$D\vec{u}f(\vec{0})=\lim_{h \rightarrow 0} \frac{f(\vec{0}+h\vec{u})-f(\vec{0})}{h}=$$

$$=\lim_{h \rightarrow 0}\frac{(hu_1,hu_2)}{h}=\lim_{h \rightarrow 0}\frac{1}{h}\frac{hu_1h^2u_2^2}{h^2u_1^2+h^4u_2^4}=$$

$$=\lim_{h \rightarrow 0}\frac{u_1u_2^2}{u_1^2}=\frac{u_2^2}{u_1}$$
si $u_1\neq0$.

En el caso en que $u_1=0$ se tiene que:

$$D\vec{u}f(\vec{0})=\lim_{h \rightarrow 0}\frac{f(0,hu_2)}{h}=\lim_{h \rightarrow 0}\frac{0}{h}=0$$

lo que prueba que $f$ tiene derivadas en cualquier dirección en el punto $\vec{x}=\vec{0}$

Obsérvese que las derivadas parciales no son continuas en este punto (en caso contrario la función hubiera sido diferenciable).

Para todos los públicos
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