Function:$$x = {- \sqrt{x^2+y^2}}.$$(a conical surface) To determine whether it has a stationary point or not, 2 condition must be met:

  • function must have partial derivatives at point p0, and
  • function must have local maxima/minima at point p0.

For this function you can calculate partial derivatives and the gradient is equal 0 when derivative with respect to x variable is equal 0: for x=0; when derivative with respect to y variable is equal 0: for y=0. Also, the cone has a MAXimum in this same point.

HOWEVER: http://www.wolframalpha.com/input/?i=-(x%5E2%2By%5E2)%5E(1%2F2)+stationary+point Wolphram hasn't found any Stationary Points, which means in point 0,0 there is no stationary point.

Doesn't the function have partial derivatives at this point? If so, why?

I tried to calculate (when I asked the question): $$\lim\limits_{x \to 0^+, y \to 0^+} - \sqrt{x^2+y^2} = 0$$ and $$\lim\limits_{x \to 0^-, y \to 0^-} - \sqrt{x^2+y^2} = 0$$

.. so I was sure the side limits are the same and equal 0. Now when I know the answer, I assume what I had done is incorrect.

function gif

wolphram alpha screen shot

  • 1
    $\begingroup$ The partial derivatives do not exist at $(0,0)$. $\endgroup$ – Paul Sep 12 '18 at 19:34
  • $\begingroup$ @Paul could you please show me or link to why doesn't partial derivatives exist? I can't see it and am bearing with it quite a time $\endgroup$ – Immo Sep 12 '18 at 19:37
  • $\begingroup$ The left-side and right-side limits are 0, no? lim x,y->0+ = lim x,y->0- $\endgroup$ – Immo Sep 12 '18 at 19:39
  • $\begingroup$ No, from one side you get $1$, and from the other side you get $-1$. $\endgroup$ – Paul Sep 12 '18 at 19:42
  • $\begingroup$ But Wolphram does calculate Derivatives wolframalpha.com/input/?i=-(x%5E2%2By%5E2)%5E(1%2F2) Also, the point (0,0) is the very top of the cone, right? So how on earth can it's limits be 1 and -1? $\endgroup$ – Immo Sep 12 '18 at 19:52

The partial derivatives don’t exist at the cone’s vertex: $f(x,0) = -(x^2)^{1/2}=-|x|$. This function doesn’t have a derivative at $x=0$, which I suspect you know. Working through the limit calculation, anyway, for $h\gt0$, the difference quotient at the origin for differentiation with respect to $x$ is $${f(h,0)-f(0,0)\over h} = -{|h|\over h}=-\frac hh=-1$$ but for $h\lt0$ it is $$-\frac{-h}h = 1,$$ therefore the limit—the partial derivative of $f$ with respect to $x$—clearly doesn’t exist. By symmetry, neither does $f_y$.

  • $\begingroup$ Thank you. Though, is "being not correct" the reason why when I calculate it in a way: $$\lim\limits_{x \to 0^+} \biggl(\lim\limits_{y \to 0^+}-\sqrt{x^2+y^2}\biggr) = 0$$ why is it wrong and I must calculate limits from definition? $\endgroup$ – Immo Sep 15 '18 at 10:08
  • $\begingroup$ @Immo As I expect you already know, the existence of a one-sided limit at a point doesn’t guarantee the existence of a limit at the point. Consider $\lim_{x\to0}\operatorname{sgn}x$, for instance. In more than one dimension, even more is required for a limit to exist. Basically, you must get the same value along all paths to the point. In your expression above, you’re only considering a subset of these paths, and only ones that lie in the first quadrant, to boot. $\endgroup$ – amd Sep 15 '18 at 19:02

lets consider a simpler example

$f(x) = - |x|$

Note that the surface of revolution of this function is a cone and is the same as the function described in the OP.

$f(x)$ has a maximum at $0$ but $f(x)$ does not have a stationary point at $0.$

$f'(0)$ does not exist.

If $x$ is a local maximum of a function either $f'(x) = 0$(and is a stationary point) or $f'(x)$ is does not exist.

Same concepts apply as you increase dimension.


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