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.


my calculations (link).


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.

  • $\begingroup$ It looks discontinuous at the origin: try taking the limit along the curves $t \mapsto (t,0)$ and $t \mapsto (t^2, t^3)$. $\endgroup$ Apr 19 '14 at 9:55
  • $\begingroup$ $$\left| \frac{x^3}{x^2+y^2} \right| \leq |x| \frac{x^2}{x^2+y^2} \to 0.$$ Hence $f$ is continuous (at zero). $\endgroup$
    – Siminore
    Apr 19 '14 at 10:02
  • $\begingroup$ In the title you talk about using definitions, then in your question you mention using "$C^1 \Rightarrow$ differentiable $\Rightarrow$ partial derivatives exists $\Rightarrow$ continous". What is it after all? $\endgroup$
    – Git Gud
    Apr 19 '14 at 10:50
  • $\begingroup$ @GitGud good question. I need to answer the question "is f contious, have partial derivatives, and is differentiable"? Definitions or not, isn't specified in the quesiton. But, after my first attempt, I see that I do need to use the definition. But as you can tell, my brain have not absorbed the theory. $\endgroup$
    – jacob
    Apr 19 '14 at 10:55
  • $\begingroup$ @jacob What is your definition of differentiable function? $\endgroup$
    – Git Gud
    Apr 19 '14 at 10:58

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

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.

Therefore $f$ is continuous.

Let us now analyze the partial derivatives.

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


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

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 $(0,0)$.

  • $\begingroup$ I have learned and used the following techniques for evaluating limits in 2 variables: * approach along different lines, say f(t,t) or f(t,0) or even f(t,t^2) etc and see if they differ. * use polar coordinates. * borin algebra tricks some of the times. question: is the absolute value-technique always/often valid (and the smart way)? I need to know this in the setting of problem solving. "Decide if lim exists or not" on an exam. $\endgroup$
    – jacob
    Apr 19 '14 at 15:24
  • 1
    $\begingroup$ I wouldn't necessarily call it the smart way, but it is useful. It is indeed valid. It's easy to replace an argument such as the one I used with an $\epsilon$-$\delta$ proof. $\endgroup$
    – Git Gud
    Apr 19 '14 at 15:28
  • $\begingroup$ I am useless at epsilon delta-proofs, can you write one (or link to some other good example, my book has to0 few)? $\endgroup$
    – jacob
    Apr 19 '14 at 15:31
  • $\begingroup$ @jacob I've added an $\varepsilon$-$\delta$ proof to the answer. Unfortunately this isn't a very instructive example. Hopefully it will still be helpful. At any rate, the absolute value boundness is enough. $\endgroup$
    – Git Gud
    Apr 19 '14 at 15:55
  • $\begingroup$ I'm lost in the details. Your answer is great; I am not. Can you tell me, either in an edit or in a comment, the headings or the "flow of logic" if you will, i.e.: first we do this, then we note this, so we have to do that, and we conclude. $\endgroup$
    – jacob
    Apr 21 '14 at 13:18

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