# Limit of Multi-Variable Functions 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.

## 1 Answer

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 $$0$$.

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