Continuity and derivability of a piecewise function $f(x,y)=\frac{x^2}{y}$

continuityderivativesmultivariable-calculuspartial derivative

Study the continuity and the derivability of the function
$$f(x,y)=\begin{cases} \frac{x^2}{y}, \ \text{if} \ y \ne 0 \\ 0, \ \text{if} \ y=0 \end{cases}$$
For the continuity: $f$ is continuous where it is defined because it is a ratio of continuous functions, the only problematic point is the junction point $(x_0,0)$ with a generic $x_0\in\mathbb{R}$.

So I have to check if
$$\lim_{(x,y)\to(x_0,0)} \frac{x^2}{y}=0$$
I've used the direction $x=\sqrt{y}+x_0$ to show that the limit doesn't exist, since for all $x_0\in\mathbb{R}$ it is
$$\lim_{(x,y)\to(x_0,0)} \frac{(\sqrt{y}+x_0)^2}{y}\ne 0$$
So $f$ is continuous for all $(x,y)\in\mathbb{R} \times \mathbb{R} \setminus \{0\}$.

For the derivability it is the same, because $f$ is a ratio of derivable functions where is is defined but, since it isn't even continuous in $(x,0)$, it can't be derivable in that point and so $f$ is derivable for all $(x,y) \in \mathbb{R} \times \mathbb{R} \setminus \{0\}$ and it is
$$f_x(x,y)= \frac{2x}{y}, \ \forall(x,y) \in \mathbb{R} \times \mathbb{R} \setminus \{0\}$$
$$f_y(x,y)=-\frac{x^2}{y^2}, \ \forall(x,y) \in \mathbb{R} \times \mathbb{R} \setminus \{0\}$$
Is this correct? I have question:

  1. in studying continuity I've chosen the direction $x=\sqrt{y}+x_0$ because I thought that the restriction must pass through the point $(x_0,0)$; is this correct or I could have used any restriction such that $y \to 0$ (for instance $x=\sqrt{y}$)?

  2. Why, for piecewise defined functions, we have to use the definition of derivability in junction points and we can't use the derivation rules? Which are the reasons? I see that we don't know if $f$ is derivable in $(x_0,0)$ yet and those rules are valid if we already know if $f$ is derivable in certain points, but I don't see from the general theory the reasons why we have to study apart the junctions points.

Thanks.

Best Answer

Actually,$$\lim_{y\to0}\frac{\left(\sqrt y+x_0\right)^2}y=\lim_{y\to0}\frac{y+2x_0\sqrt y+x_0^{\,2}}y=\lim_{y\to0}1+2\frac{x_0}{\sqrt y}+\frac{x_0^{\,2}}y,$$and this limit doesn't exist (in $\Bbb R$). So, $f$ is discontinuous at those points. In particular, it is not differentiable there.

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