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Evaluate $$\lim_{(x,y)\to (0,0)}\frac{\sin(xy^2)}{x}$$

$$\lim_{(x,y)\to (0,0)}\frac{\sin(xy^2)}{x}=\lim_{(x,y)\to (0,0)}\frac{\sin(xy^2)}{xy^2}y^2$$

now $$\lim_{(x,y)\to (0,0)}\frac{\sin(xy^2)}{xy^2}\cdot \lim_{(x,y)\to (0,0)} y^2=1\cdot 0=0$$

  1. Is there a problem with this calculation regarding the where the function is defined?

  2. In general if we look at $(x,0)\to(0,0)$ or $(0,y)\to(0,0)$ are those iterative limits?

Botond
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gbox
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  • What does your 2nd question mean? – Botond Mar 06 '18 at 13:41
  • @Botond if we look at $lim_{(x,0)\to (0,0)}f(x)$ or $lim_{(0,y)\to (0,0)}f(x)$ is it the same as looking at $lim_{x \to 0}(lim_{y \to 0}f(x))$ or $lim_{y\to 0}(lim_{x \to 0}f(x))$ respectively – gbox Mar 06 '18 at 13:48
  • IFAIK it is not the same, hope somebody will provide a good example for it – mike239x Mar 06 '18 at 13:50
  • @gbox What do you mean by $(x,0) \to (0,0)$? You'd like to substitute $y=0$ and then take $x\to0$? – Botond Mar 06 '18 at 13:51
  • @gbox ah ok you are asking for the case x=0 and $y\neq 0$? – user Mar 06 '18 at 13:51
  • @gimusi shouldn't it be obvious? (not asking in disrespect) if we take a limit $(x,y)\to (0,0)$ $x,y$ can be defined/undefined in $0$ no? (Removable discontinuity) – gbox Mar 06 '18 at 13:55
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    @gbox we can't take the limit for $(0,y)\to (0,0)$ thus the limit does not exist unless we remove $x=0$. – user Mar 06 '18 at 13:56
  • @gimusi I see, and in the case, and in general, looking at $(0,y)\to (0,0)$ is the same as looking at $lim_{y\to 0}(lim_{x\to 0}f(x))$? I think that this case show that it is not – gbox Mar 06 '18 at 14:03
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    @gbox since f(x,y) is not defined for $x=0$ the limit does not exist (not $x\to0$ but $x=0$ that is y axis), if we remove this discontinuity the limit exists – user Mar 06 '18 at 14:05

1 Answers1

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Note that, since for $x=0$ the expression is not defined, the limit does not exist.

If we define $f(x,y)=0$ for $x=0$ then by squeeze theorem

$$0<\frac{\sin(xy^2)}{x}=\frac{\sin(xy^2)}{xy^2}y^2<y^2\to 0$$

the limit exists and is equal to $0$.

user
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  • So we should not look at $\lim_{(x,y)\to (0,0)}\frac{\sin(xy^2)}{xy^2}\cdot \lim_{(x,y)\to (0,0)} y^2=1\cdot 0=0$ , but rather by the squeeze theorem? – gbox Mar 06 '18 at 13:50