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For the function,

$$ \begin{aligned} f(x,y) = \begin{cases} \hfill 0 \hfill &,\mbox{ if } x = y = 0 \\ \hfill \frac{x^{2}y}{x^{4}+y^{2}} \hfill &, \mbox{ otherwise} \end{cases} \end{aligned} $$

Show it is discontinuous at $x=y=0$ by taking the limit $x \to 0$ along the line $y = {x^2}$. I don't even know what this means, I have a really unclear professor who mumbles through things half the time. Any assistance is appreciated.

EDIT:

What about the function, $$g(x,y) = \left\{ \matrix{ 0,x = y = 0 \hfill \cr {{{x^2}y} \over {{x^4} + {y^2} + 1}},otherwise \hfill \cr} \right\}$$

How do I know whether it's continuous at x=y=0 because I can't make a substitution like before...

Ahmed
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1 Answers1

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If a function of two variables should be continuos, that is $\lim_{(x,y) \to (a,b)} f(x,y) = f(a,b)$ independent on how you approach $a,b$. You could think of this like that you are drawing a circle around the point $(a,b)$ and independent from which point on the circle that you start from "walking" to the point $(a,b)$ you should "end" at the same value.

So if we choose to start from the point $(x,x^2)$ we should end up at the value $0$ IF is continouis.

$$ f(x,x^2) = \frac{x^4}{x^4 + x^4} = \frac{1}{2} \not = 0, \ x \not = 0 $$

According to your second question, remember from early mathematics when you first learned about limits. I think it went something like this.

Lets say you have a one-dimension function $f(x)$ and you wanted to calculate the limit when $x \to a$, the first thing you would do is try to calculate $f(a)$. It $f(a)$ was definied, then you would simply say that $\lim_{x \to a} f(x) = f(a)$. If $f(a)$ was not definied, then you would have to do some "tricks".

In this new case one simply sees that $$ f(0,0) = \frac{0}{1} = 0 $$ So this is no "dangerous" case, so we conlcude that $\lim_{(x,y) \to (0,0)} f(x,y) = f(0,0)$

Olba12
  • 2,579
  • What about the function $$g(x,y) = \left{ \matrix{ 0,x = y = 0 \hfill \cr {{{x^2}y} \over {{x^4} + {y^2} + 1}},otherwise \hfill \cr} \right}$$ How do I know if that is continuous at x=y=0? With this problem and a few others, I've had a substitution to make which makes it easy now that you've shown me how to solve this type of problem. – Ahmed May 11 '16 at 20:03
  • I think the best is that you make a new question (according to the rules) or that you add the above to this question using edit. Then I will be happy to answer. @Ahmed – Olba12 May 11 '16 at 20:17
  • Oh, ok. I am very sorry! I will add it to the question now. – Ahmed May 11 '16 at 21:52
  • For some reason, the edit didn't work before. I am not sure what I did wrong but I have now included it in the question. Thanks again. – Ahmed May 12 '16 at 11:36
  • I have editet @Ahmed – Olba12 May 12 '16 at 11:55