Let $f: \mathbb{R}^2 \to \mathbb{R}$ be defined by:

$f(x,y) = \begin{cases}y - x^2, & y\geq x^2 \\0, & -x^2\leq y < x^2 \\ y + x^2, & y < -x^2\end{cases}$

(a) Is $f$ differentiable at the origin?

(b) Is f of class $C^1$ at the origin?

My attempt at a solution:

$\frac{\partial f}{\partial x}(0,0)$ = $\lim_{h\to0}\frac{f(h,0) - f(0,0)}{h} =$ $\lim_{h\to0}\frac{(0 - h^2) - (0-0)}{h} = -h^2/h =$$-h = 0$

$\frac{\partial f}{\partial y}(0,0)$ = $\lim_{h\to0}\frac{f(0,h) - f(0,0)}{h} =$ $\lim_{h\to0}\frac{(h - 0) - (0-0)}{h} = h/h = 1$

So the partial derivatives exist at $(0,0)$ and $\nabla f = (0,1)$

(b) I can see that $f$ is not $C^1$ since the partial derivatives of $f$ with respect to $y$ are not continuous at $(0,0)$

(a) If $f$ were differentiable at the origin, then:

$\lim_{(s,t) \to (0,0)} \frac{f(0+s,0+t) - f(0,0) - \langle 0,1 \rangle \cdot (s,t)}{|(s,t)|} \to 0$

which simplifies to:

$\lim_{(s,t) \to (0,0)} \frac{f(s,t)-t}{|(s,t)|} = 0$

If this were true then:

$(1) \qquad \lim_{(s,t) \to (0,0)} \frac{-s^2}{\sqrt{s^2+t^2}} = 0, \qquad \text{on the set where $t \ge s^2$}$

$(2) \qquad \lim_{(s,t) \to (0,0)} \frac{-t}{\sqrt{s^2+t^2}} = 0, \qquad \text{on the set where $-s^2 \leq t < s^2$}$

$(3) \qquad \lim_{(s,t) \to (0,0)} \frac{s^2}{\sqrt{s^2+t^2}} = 0, \qquad \text{on the set where $t < - s^2$}$

Approaching along the linear path $s = mt$ we see that $(1)$ and $(3)$ evaluate to $0$ but that $(2)$ evaluates to $\frac{-1}{\sqrt{m+1}}$

Thus the limit

$\lim_{(s,t) \to (0,0)} \frac{f(s,t)-t}{|(s,t)|}$ does not exist and $f$ is not differentiable at the origin.

Is this a correct solution?

The only other way I can think to do this question is by using the properties of direction derivatives.

Let $u = (sin (\theta), cos(\theta))$

Then the directional derivative of $f$ at $(0,0)$ in the direction of $u$ is defined to be

$\partial_u f(0,0)= lim_{t \to 0} \frac{f((0,0) +t(sin(\theta), cos(\theta)) - f(0,0)}{t}$

$= lim_{t \to 0} \frac{f(tsin(\theta), tcos(\theta))}{t}$

$= lim_{t \to 0} \frac{tcos(\theta) -t^2sin^2(\theta)}{t}$

$= cos(\theta)$

So the directional derivatives of $f(0,0)$ all exist and are given by:

$\partial f(0,0) = \nabla f(0,0)\bullet u = (0,1) \bullet (sin(\theta), cos(\theta)) =(0,cos(\theta))$

Wouldn't this imply that the f is differentiable at the origin?

Furthermore, since all of the partial derivatives at the origin exist and are continuous, $f$ is $C^1$ at the origin?

  • $\begingroup$ Your evaluation of (2) is not correct, because the linear path $s=mt$ is not contained in the region $|t| < s^2$: the inequality $$|t| < m^2 t^2$$ is equivalent to $$1 < m^2 |t|$$ which is false for $t$ near zero. So you have to do something else to evaluate (2). $\endgroup$
    – Lee Mosher
    Apr 21, 2018 at 12:41
  • $\begingroup$ @LeeMosher for (2) could I use the path $s = 100\sqrt t$? $\endgroup$ Apr 21, 2018 at 21:13
  • $\begingroup$ @LeeMosher But then (2) would evaluate to $0$ and this path doesn't satisfy the conditions for (1) or (3). Do I need to find a path the satisfies the conditions for all (1), (2) and (3)? And if so, what do I do if they all evaluate to the same thing? This just shows the limits exists along one path right? I'm really confused. $\endgroup$ Apr 21, 2018 at 21:22
  • $\begingroup$ I added more to my answer. $\endgroup$
    – Lee Mosher
    Apr 21, 2018 at 21:51
  • $\begingroup$ @LeeMosher Ohhh, I see now. Thank you. This really clears things up. I also updated my attempted solution with an argument that uses directional derivatives. I'm not sure if it's a valid argument or not but maybe you could take a look at it. $\endgroup$ Apr 21, 2018 at 22:12

1 Answer 1


Certainly $f$ is not of class $C^1$ at the origin: there is no neighborhood of $(0,0)$ in which $\partial f/\partial y$ exists and is continuous. To see why, in the region $-x^2 < y < x^2$ we have $\partial f/\partial y = 1$, whereas in the region $y \ge x^2$ we have $\partial f/\partial y = 0$. These do not match up continuously at any point of the parabola $y=x^2$, and yet every neighborhood of $(0,0)$ has points on that parabola.

Also, in your attempt to prove that $f$ is differentiable, perhaps you wanted to use the theorem which says that if $f$ is $C^1$ in a neighborhood of $(0,0)$ then $f$ is differentiable in that neighborhood. But, since $f$ is not $C^1$ in any neighborhood, this theorem gives no information about differentiability at $(0,0)$.

So it looks like you have to go back to the definition of the multivariable derivative. You already know that $\partial f/\partial x(0,0)=0$ and $\partial f/\partial x(0,0)=1$. So what you have to determine is whether $$\lim_{(s,t) \to (0,0)} \frac{f(0+s,0+t) - f(0,0) - \langle 0,1 \rangle \cdot (s,t)}{|(s,t)|} \to 0 $$ which simplifies to $$\lim_{(s,t) \to (0,0)} \frac{f(s,t)-t}{|(s,t)|} = 0 $$ This breaks into three separate limits, and you have to check whether each is true: $$(1) \qquad \lim_{(s,t) \to (0,0)} \frac{-s^2}{\sqrt{s^2+t^2}} = 0, \qquad \text{on the set where $t \ge s^2$} $$ $$(2) \qquad \lim_{(s,t) \to (0,0)} \frac{-t}{\sqrt{s^2+t^2}} = 0, \qquad \text{on the set where $-s^2 < t < s^2$} $$ $$(3) \qquad \lim_{(s,t) \to (0,0)} \frac{s^2}{\sqrt{s^2+t^2}} = 0, \qquad \text{on the set where $t \le - s^2$} $$ Are they true?

It's pretty easy to see that (1) is true, because the limit equation is true on the whole $(s,t)$ plane, since $f(x,y)=y-x^2$ is differentiable with partial derivatives equal to $0$ at $(0,0)$. So the limit is true when restricted to $t \ge s^2$. Similarly for (3).

The limit (2) is also true, although the proof is a bit trickier. If we divide the top and bottom by $s$ we'll get $$\frac{-\frac{t}{s}}{\sqrt{1 + \bigl(\frac{t}{s}\bigr)^2}} $$ As $(s,t)$ approaches $(0,0)$ in the region $-s^2 \le t < s^2$, the ratio $\frac{t}{s}$ approaches $0$. So the limiting value is $$\frac{-0}{\sqrt{1 + \bigl(0\bigr)^2}} = 0 $$

Notice, these limit arguments are not about "approaching $(0,0)$ along paths". Thinking about approaching $(0,0)$ along paths can be a useful argument for disproving a limit equation, but for proving a limit equation it's not a very useful way to think.

  • $\begingroup$ I updated my solution. Is it correct? $\endgroup$ Apr 20, 2018 at 23:17
  • $\begingroup$ @Lee_Mosher I updated my solution, Is it correct? $\endgroup$ Apr 21, 2018 at 0:12
  • $\begingroup$ No, your evaluation of (2) is not correct. See my comment to our answer. $\endgroup$
    – Lee Mosher
    Apr 21, 2018 at 12:41

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