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I need help finding if the limit of $g(x,y)$ at $(0,0)$ and at $(2,0)$ to see if it is continuous on these points.

I get confused, because normally these piecewise functions are defined such that $(x,y)$ different than some $(a,b)$ but here I have no idea how to do it because I can approach $(0,0)$ or $(2,0)$ from both pieces of the function...

And if I wanted to see the differentiability as well, what should I do?

I would really appreciate if you could explain this to me. Thank you !

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$g(x,y)$ is continuous at $(x_0,y_0)$ if $\forall \epsilon > 0$, $\exists \delta > 0$ $|| (x,y) - (x_0,y_0)|| < \delta \Rightarrow || g(x,y) - g(x_0,y_0)|| < \epsilon$.

Continuity at $(0,0)$: Let $\epsilon > 0$. Let $\delta = \min(\frac{\sqrt{\epsilon}}{2}, \frac{1}{2})$. \begin{align*} || (x,y) - (0,0)|| < \delta &\Rightarrow |x| < \delta \land |y| < \delta \\ \Rightarrow || g(x,y) - g(0,0)|| &= |xy + xy^3 \sin(\frac{x}{y})| \\ &\leq |xy| + |xy^3| < \delta^2 + \delta^4 \leq \epsilon \: (\text{check}!) \end{align*} Hence g is continuous at $(0,0)$. You could check for $(2,0)$.

$g(x,y)$ is differentiable at $(x_0,y_0)$ if $\exists (\alpha, \beta)$ s.t. $\epsilon(h_1, h_2) := \frac{g(x_0 + h_1, x_0 + h_2) - g(x_0,y_0) - (\alpha, \beta) \odot (h_1, h_2)}{||(h_1,h_2)||} \rightarrow 0$ as $(h_1, h_2) \rightarrow 0$. (where $\odot$ means dot product.)

You could check differentiability using this definition.

If you haven't seen these definitions before, then what you want to search for is 'continuity' and 'differentiability' of "multi-variable functions". Also there is another definition of continuity using sequences instead of $\epsilon-\delta$, you could again search for that (on google, or any calculus book on vector valued functions).

Edit (after seeing comment):

Intuitive understanding of continuity:

Continuity of $\mathbf{f: \mathbb{R} \rightarrow \mathbb{R}}$ at $\mathbf{x_0 \in \mathbb{R}}$.

Visualize $x_0$ on the real number line. The definition of continuity would mean "if you approach $x_0$ from any side, then it's corresponding value of $f(x)$ must approach $f(x_0)$. Note that since x is a real number, you can approach it from two sides - left and right leading to the definition of left hand limits and right hand limits etc.

Continuity of $\mathbf{f: \mathbb{R}^2 \rightarrow \mathbb{R}}$ at $\mathbf{(x_0,y_0) \in \mathbb{R}^2}$.

Visualize $(x_0, y_0)$ on the 2D plane (x-y plane). The definition of continuity would mean "if you approach $(x_0, y_0)$ from any side, then it's corresponding value of $f(x,y)$ must approach $f(x_0, y_0)$. Now however as $(x_0,y_0)$ is a point on the 2D plane, you can approach it from infinitely different directions. For example, you could approach this point via a line segment ending at $(x_0,y_0)$ with a slope of $k$ for any value $k$. (This actually leads to the definition of the $k$^th directional derivative). Alternatively you could approach it along any path which finally terminates at $(x_0,y_0)$ like a spiral etc.

If this is a bit hard to understand, let me know, I'll include some figures.

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  • $\begingroup$ Ohhh okay! Thanks a lot! I will try to learn this. So for this kind of limit, it would be wrong to use polar coordinates to show that the limit when (x,y) ---> (0,0) of (xy +xy^3*sin(x/y)) and that the limit as (x,y) --> (0,0) of (0) are both zero. And then concluding that because the limit as (x,y)--> (0,0) of both parts is zero then the limit exists and it is zero. Would this be wrong??? This is they only way I could think about before I asked, but I think it might not be correct, and your way is much better. Thank you! $\endgroup$ – Gema Mar 20 at 21:50
  • $\begingroup$ That depends on your understanding of "(x,y) --> (0,0)". I'll edit my answer for some explanation. $\endgroup$ – Kaind Mar 20 at 21:54
  • $\begingroup$ And for the differentiability. How do I know what to put for 𝑔(𝑥0+ℎ1,𝑥0+ℎ2) . Because I get confused with this, so I don't know if 𝑔(𝑥0+ℎ1,𝑥0+ℎ2) = 0 or 𝑔(𝑥0+ℎ1,𝑥0+ℎ2) = h1*h2 + h1*h2^3*sin(h1/h2) $\endgroup$ – Gema Mar 20 at 21:55
  • $\begingroup$ Thanks a lot for everything! $\endgroup$ – Gema Mar 20 at 21:55
  • $\begingroup$ For the differentiability bit, if you are computing the derivative of $(0,0)$, then you have $g(h_1, h_2)$, and since $(h_1,h_2) \rightarrow (0,0) \Rightarrow (h_1,h_2) \neq (0,0)$. So you let $g(h_1,h_2) = h_1h_2 + h_1h_2^3\sin( \frac{h_2}{ h_2})$. Also note that in the differentiability bit, $(\alpha, \beta)$ is called the derivative of $g$ at $(x_0,y_0)$. And a guess candidate for $(\alpha, \beta)$ which will work if $g$ is differentiable is $(\alpha, \beta) = (\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y})$ (partial derivatives). $\endgroup$ – Kaind Mar 20 at 22:10
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For $\vert y \vert \le 1$ and $y \neq 0$, you have

$$\vert g(x,y) \vert \le \vert x y \vert + \vert x \vert \vert y \vert^3 \le 2 \vert x \vert \vert y \vert.$$

As the right hand side of this inequality goes to zero as $y \to 0$ and $g(x,0)=0$ for all $x$, we get that $g$ is continuous everywhere.

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  • $\begingroup$ Thanks! But why is the absolute value of y less than one. How do you know that? Couldn't y = 2 for example? Thank you! $\endgroup$ – Gema Mar 20 at 21:51
  • $\begingroup$ The absolute value is not less than one in general. But when you are looking to the limit $y \to 0$, you can limit your analysis to $\vert y \vert \le1$. $\endgroup$ – mathcounterexamples.net Mar 21 at 8:33

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