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Question: Given $f:\mathbb{R}^2 \to \mathbb{R}$ defined by $f(x, y) = \begin{cases} x, & \text{if $y=x^2$} \\ 0, & \text{otherwise} \end{cases}$, show $f$ is continuous at $(0, 0)$.

Attempt: I have proven that $f$ is not differentiable at $(0, 0)$, which means I can't imply continuity from it being differentiable.

I know that I must show $\lim_{(x,y) \to (0, 0)} f = f(0, 0)=0$, but am not experienced in proving the limits of multivariable functions (at least not for ones that you can't apply the squeeze theorem to).

Or is there a way to utilise the continuity of the directional derivatives?

I know this is a pretty weak attempt but I am really stuck, and any help would be greatly appreciated.

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    $\begingroup$ Do you really need derivatives though? Can't you apply squeeze theorem with $g(x,y)=|x|$ and $h(x,y)=-|x|$ $\endgroup$
    – Anvit
    Jun 1, 2020 at 11:46

2 Answers 2

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When dealing with multivariate functions like this one, you consider $(x,y)\to (0,0)$.

This means $x\to 0$ and $y\to 0$.

  • Most of the time you want to show $|f(x,y)-\ell|<k\ |x|^\alpha|y|^\beta\to 0$ for some $\alpha,\beta > 0$, since $|x|^\alpha\to 0$ and $|y|^\beta\to 0$.

Note that it is not mandatory to have both $x$ and $y$ on the RHS (i.e. $\alpha=0$ or $\beta=0$ is ok).

You can of course also have some other combination of $|x|^\alpha$ and $|y|^\beta$ (a sum for instance).

  • In some cases using polar coordinates is helpful, you get $|f(x,y)-\ell|<\phi(\theta)\ |r|^\gamma\to 0\ \ $ provided $|\phi(\theta)|<M$ is bounded.

Since $r^2=x^2+y^2\to 0$ then $|r|\to 0$.

  • Another technique is to set $t(x,y)=\frac yx$ or $y=tx$ to get $|f(x,y)-\ell|<\phi(t)|x|^\alpha\to 0\ $ provided again that $|\phi(t)|<M$ is bounded.

In some cases, $\alpha=0$ and you need to show that $\phi$ itself tends to zero.


In our case, we are in the best conditions possible since $\ell=0$ and $|f(x,y)|\le|x|\to 0$.

So this is over and you can claim $\lim\limits_{(x,y)\to (0,0)}f(x,y)=0$, making $f$ continuous in $(0,0)$.

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Attempt:

Let $\epsilon >0$ be given.

Need to show that for $\epsilon >0$ there is a $\delta$ s.t.

$\sqrt{x^2+y^2}<\delta$ implies

$|f(x,y)-0|<\epsilon$.

Choose $\delta=\epsilon$.

Then

$0 \le |f(x,y)-0| \le |x|=√x^2 \le$

$\sqrt{x^2+y^2} <\delta =\epsilon$.

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  • $\begingroup$ May I ask where you got the $\sqrt{x^2 + y^2}$? $\endgroup$
    – Viv4660
    Jun 1, 2020 at 12:12
  • $\begingroup$ He is using $||(x,y)||_2$ the euclidean norm, but in $\mathbb R^n$ all norms are equivalent, so it doesn't really matter, you choose $1-$norm $|x|+|y|\to 0$ or $\infty-$norm $\max(|x|,|y|)\to 0$ or $2-$norm $|r|\to 0$. $\endgroup$
    – zwim
    Jun 1, 2020 at 12:16
  • $\begingroup$ Sure. $\sqrt{x^2} \le \sqrt{x^2+y^2}$, added $y^2 \ge 0$, inequality is ok. Want to get an upper bound for $|f(x,y)|$ that relates $\delta$ and $\epsilon$.Your thoughts? $\endgroup$ Jun 1, 2020 at 12:16

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