# Proving continuity of $f:\mathbb{R}^2 \to \mathbb{R}$ at $(0, 0)$

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.

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

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 $$|f(x,y)-\ell| for some $$\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)| 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)| 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 $$(0,0)$$.

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

$$0 \le |f(x,y)-0| \le |x|=√x^2 \le$$

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

• May I ask where you got the $\sqrt{x^2 + y^2}$? Jun 1, 2020 at 12:12
• 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 $|r|\to 0$.
– zwim
Jun 1, 2020 at 12:16
• 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? Jun 1, 2020 at 12:16