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I try taking the limit along $ y = 0 $ and get $ 4x\sin \frac{1}{x} $ but when I try substituting the value x=0 and/or solving the limit along $ x=0 $, the function is undefined so I'm confused how to approach this. I'm assuming the function is also not continuous at $ (0,0) $ since the first piece wise is undefined for $ x=0 $ and that function differs from $ e^{-5y^3} $. I'm just not sure how to approach this.

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The limit along $x=0$ is $1$ whereas $|f(x,y)| \leq 4|x|e^{2} (\to 0)$ whenever $\|x|$ and $|y|$ are sufficiently small so the limit does not exist and $f$ is not continuous.

More explicitly consider $\lim_{n \to \infty} f(0,\frac 1 n)$ and $\lim_{n \to \infty} f(\frac 1 n,0)$. The first limit obviously $1$. For the second limit use the fact that $|sin(\frac 1 x)|\leq 1$. You can see that the limit is $0$.

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  • $\begingroup$ How did you conclude this? $\endgroup$ – Hopper Feb 28 '19 at 7:15
  • $\begingroup$ @krauser126 See my edited answer. $\endgroup$ – Kavi Rama Murthy Feb 28 '19 at 7:27

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