0
$\begingroup$

Let $$(f :\mathbb{R}^2 \to \mathbb{R})$$ be defined by $$\frac{x_1x_2^2}{(x_1^2+x_2^2)}$$ if $x$ $\neq(0,0)$, $f(0)$ = $0$.

Show that $f$ is continuous at $(0, 0)$.

Show that the partial derivatives $\frac{∂f}{∂x1}(0, 0),$ $\frac{∂f}{∂x2} (0, 0)$ exist and find their values.

(c) By using the definition ∗ show that f is not differentiable at $x$ = $0$.

Note that $0$ and $x$ are vectors

The * from above : $\lim_{x\to a} \frac{||F(x) - F(a)-DF(a)(x-a)||}{||x-a||} = 0.$

For a) I know the limit is 0, but wouldn't that mean the function is not continuous at 0? Or is it just 0? For b) if I take the partial's won't I just end up with 0? Not sure how to do c.

Any help is appreciated!

  • 0
    For (a), you are correct that the limit as $(x_1,x_2)\rightarrow (0,0)$ is zero (but have you proven it?). Since the limit is zero and $f(0,0) = 0,$ this establishes that the function is continuous at $(0,0).$ Recall that the definition of continuity is that the limit equals the function value. For (b), I agree that both partial derivatives are zero. For (c) I am unfamiliar with the notation $DF(a)(x-a).$ However, it is not differentiable because the directional derivative along, say a 45 degree angle is undefined, unlike in the $x$/$y$ directions where in (b) you showed it exists and is 0. – 2017-01-26
  • 0
    I used squeeze theorem to prove that the function's limit is $0$. So using that limit and the fact that the function is defined at 0, I can conclude it is continuous? For c) $DF(a)(x-a)$ the $DF(a)$ part is a matrix of partial derivatives according to my textbook. – 2017-01-26
  • 0
    Using that and the fact that the function is defined *to be $0$* at $0$ you can conclude that. I'd like an explicit definition, but would it suffice to say that term is zero since the partial derivatives are both zero? – 2017-01-26
  • 0
    But can you conclude an answer from part b) in part a? – 2017-01-26
  • 0
    Huh? what answer from what? – 2017-01-26
  • 0
    Also, I was wrong when I said why the function was not differentiable in my first comment (all the directional derivatives do, in fact, exist, so we need to fall back to the definition you were given... see my answer below) – 2017-01-26
  • 0
    I was asking if it is okay to use what I know from part b, to answer the question for part a. ie. using the fact that the partial derivatives are 0 to conclude that the function is continuous at 0. Normally in questions with multiple parts, you wouldn't use a fact from a later part to answer a previous part. And yes, the explanation below makes perfect sense! – 2017-01-26
  • 0
    So while proving the limits of the partials, I ended up getting non zero values, I ended up with the $\lim_{(x)\rightarrow(0}\frac{h}{h}$ which is 1 and the limit would be 1 as well. Where did I go wrong? Similar situation for y. – 2017-01-26
  • 0
    No, the existence of partial derivatives (or even of all directional derivatives) does not imply continuity. As a counterexample $f(x,y) = xy/(x^2+y^2)$ works. Differentiability does imply continuity. – 2017-01-26
  • 0
    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52507/discussion-between-spaceisdarkgreen-and-quwerty). – 2017-01-26

1 Answers 1

1

For part (c) since the partial derivatives are all zero, the formula reduces to proving that $$\lim_{(x,y)\rightarrow(0,0)}\frac{|f(x,y)|}{\sqrt{x^2+y^2}}\ne 0$$ Going to polar cooridinates we can write the expression as $$ \frac{r\cos(\theta)r^2\sin^2(\theta)}{r^3}=\cos(\theta)\sin^2(\theta)$$ So we see that the limit is not zero in any direction except along the $x$ or $y$ axis, and thus the limit does not exist.