Find out if $f$ differentiable, continuous and have partial derivatives - or if it does not. Using definitions.

Question

Problem

Find out if the function

$$f(x)=\left\{\begin{array}{l} \frac{x^3}{x^2+y^2},\:\text{if $(x,y) \not= (0,0)$;}\\ 0,\:\text{if $(x,y)=(0,0)$;} \end{array}\right.$$

is continuous, have partial derivatives, and is differentiable.

Attempt

my calculations (link).

Question

Specific questions: See the three yellow places in my calculations.

General question: I tried doing this via:

$C^1 \Rightarrow$ differentiable $\Rightarrow$ partial derivatives exists $\Rightarrow$ continuous. Starting with $C^1$ (since the test is easy) and then moving on the the other cases, using the definitions.

Answer 1

The only problematic point is $(0,0)$.

As Siminore noted in the comments, for all $(x,y)\in \mathbb R^2\setminus \{(0,0)\}$ it holds that

$$|f(x,y)|\leq |x|\dfrac {x^2}{x^2+y^2}\leq |x|.$$

Consequently $$\lim \limits_{(x,y)\to (0,0)}\left(|f(x,y)|\right)\leq \lim \limits_{(x,y)\to (0,0)}\left(|x|\right)=0,$$

which implies $\lim \limits_{(x,y)\to (0,0)}\left(f(x,y)\right)=0=f(0,0)$.

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For an $\varepsilon$-$\delta$ argument recall that $\lim \limits_{(x,y)\to (0,0)}\left(f(x,y)\right)$ if, and only if, $$\forall \varepsilon >0\,\exists \delta >0\, \forall (x,y)\in \mathbb R^2\left(\Vert (x,y)-(0,0)\Vert<\delta \implies |f(x,y)-f(0,0)|<\varepsilon\right).$$

Take $\varepsilon >0$, let $\delta=\epsilon$ and let $(x,y)\in \mathbb R^2\setminus \{(0,0)\}$. The case $(x,y)=(0,0)$ is logically trivial.

You wish to prove that $\Vert (x,y)\Vert<\delta \implies \left|\dfrac {x^3}{x^2+y^2}\right|<\varepsilon$.

Assume $\Vert (x,y)\Vert<\delta$.Then $|x|\leq \sqrt{x^2+y^2}=\Vert (x,y)\Vert<\delta$ follows.

But since $|f(x,y)|\leq |x|\dfrac {x^2}{x^2+y^2}\leq |x|$, it follows that $|f(x,y)|<\delta=\varepsilon$, as wished.

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Therefore $f$ is continuous.

Let us now analyze the partial derivatives.

One has $$\begin{cases} \dfrac {\partial f}{\partial x}(x,y)= \dfrac{x^4+3x^2y^2}{(x^2+y^2)^2}\\ \dfrac {\partial f}{\partial x}(0,0)=\lim \limits_{h\to 0}\left(\dfrac {f(h,0)}{h}\right)=\lim \limits_{h\to 0}\left(\dfrac {h^3/h} h\right)=1\end{cases}$$

and

$$\begin{cases} \dfrac {\partial f}{\partial y}(x,y)= \dfrac{-2x^3y}{(x^2+y^2)^2}\\ \dfrac {\partial f}{\partial y}(0,0)=\lim \limits_{h\to 0}\left(\dfrac {f(0,h)}{h}\right)=\lim \limits_{h\to 0}\left(\dfrac 0 h\right)=0\end{cases}$$

So the partial derivatives exist in $\mathbb R^2$, (even though the partial derivative with respect to $y$ isn't continuous - this isn't enough for non-differentiability).

One has, (as a quick consequence of the definition), that $f$ is differentiable at $(0,0)$ if, and only if the limit below exists:

$$\lim_{(x,y) \to (0,0)} \left[\dfrac{f(x,y) - \left(f(0,0) + f_{x}(0,0)(x-0) + f_{y}(0,0)(y-0)\right)}{\sqrt{x^2 + y^2}}\right].$$

This limit, if it exists, equals $\lim \limits_{(x,y)\to (0,0)}\left(-\dfrac{xy^2}{\left(x^2+y^2\right)^{3/2}}\right).$

But, given $k\in \mathbb R$, $$\lim \limits_{\substack{(x,y)\to (0,0)\\x=ky}}\left(-\dfrac{xy^2}{x^2+y^2}\right)=\lim \limits_{\substack{(x,y)\to (0,0)\\x=ky}}\left(-\dfrac{ky^3}{\left(k^2y^2+y^2\right)^{3/2}}\right)=\dfrac {k}{\left(k^2+1\right)^{3/2}},$$

so the limit is path dependent and it doesn't exist.

Therefore $f$ isn't differentiable at $(0,0)$.