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show that limit doesn't exist for $f: \mathbb{R^2} \backslash \{(0,0) \} \to \mathbb{R}$ $$ f(x,y) = \frac{x^3-y^3}{x^3 + y^3} $$

would I be right in thinking I can choose points? Like say I wrtie:

let $x = 0$ and let $y \to 0$ and we get $-1$ as the limit, and similary letting $y = 0$ and $x \to 0$ then the limit would be $1$? is that all that is required?

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    $\begingroup$ I'm not entirely convinced your function is well-defined on $\mathbb{R}^2\setminus\{(0,0)\}$. $\endgroup$ – Clement C. Oct 9 '15 at 21:11
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    $\begingroup$ Unless you are specifically asked to give a formal $\epsilon$-$\delta$ argument, finding two paths along which the limits are different is enough. $\endgroup$ – André Nicolas Oct 9 '15 at 21:12
  • $\begingroup$ @ClementC. I just made up the function so it may not be, I was just worried about the general method $\endgroup$ – FACEIT Oct 9 '15 at 21:12
  • $\begingroup$ You are right, that's all. $\endgroup$ – dafinguzman Oct 9 '15 at 21:12
  • $\begingroup$ general method is taking paths. $\endgroup$ – Luis Felipe Oct 9 '15 at 21:12
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if we take the path $(t,t)$ then the function goes to $0$. but if we take the path $(2t,t)$, the function goes to $7\over 3$. so, limit doesn't exists in $0$.

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  • $\begingroup$ Is there any thing wrong with my path? $\endgroup$ – FACEIT Oct 9 '15 at 21:15
  • $\begingroup$ No, nothing. Your idea works. $\endgroup$ – Clement C. Oct 9 '15 at 21:15
  • $\begingroup$ Your idea is ok $\endgroup$ – Luis Felipe Oct 9 '15 at 21:18

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