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Question Number 176652 by floor(10²Eta[1]) last updated on 24/Sep/22

$$\mathrm{f}(\mathrm{x},\mathrm{y})=\{\begin{array}{l}{\mathrm{e}}^{\frac{1}{{\mathrm{r}}^2-1}}\mathrm{if}\mathrm{r}<1,\mathrm{where}\mathrm{r}=\parallel (\mathrm{x},\mathrm{y})\parallel \\ 0\mathrm{if}\mathrm{r}⩾1\end{array}$$ $$\mathrm{show}\mathrm{that}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{is}\mathrm{continuous}\mathrm{in}{\mathbb{R}}^2$$

Answered by floor(10²Eta[1]) last updated on 24/Sep/22

$$\mathrm{recall}\mathrm{that}\parallel (\mathrm{x},\mathrm{y})\parallel =\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}$$

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