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Determine if the following function is continuous in $(0,0)$. $$f: \mathbb{R}^2 \rightarrow \mathbb{R},\begin{pmatrix}x\\y\\\end{pmatrix} \rightarrow \begin{cases} 1& ,x≤ 0, y \in \mathbb{R} \\ x^{y^2}& , x > 0, y\in \mathbb{R} \end{cases}$$

I think $f$ is not continuous. Using that $x^{y^2} = \exp(\ln(x) \cdot y^2)$ I thought I could find a path to show that, but I don't see it.

I'd appreciate any help.

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I think you have your piece wise conditions backwards. – symplectomorphic Jun 1 '14 at 3:13
@symplectomorphic I corrected a few errors, sorry. – Nhat Jun 1 '14 at 3:16
Is $y$ really just a number in $\mathbb{R}$? I'm not sure why you don't just say that $y$ is fixed then. – Clarinetist Jun 1 '14 at 3:22
I am likewise unsure. I'm going to grab my Real Analysis text and see if I can figure out a way to do this one... – Clarinetist Jun 1 '14 at 3:46
Try something like $x = e^{-1/y^2}$ – Carl Jun 1 '14 at 4:10
up vote 1 down vote accepted

Consider the path $x=e^{-1/y^{2}}$ as $y\rightarrow 0$.

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