> [!NOTE] > A function that [[Separately Continuous Functions|separately continuous]] may not [[Continuous maps|continuous]]. ###### Proof by counter-example Define $f:\mathbb{R}^2 \to \mathbb{R}$ by $f(x,y):=1, \quad \text{if }xy \neq 0, \quad \quad \quad f(x,y):= 0 , \quad \text{if }xy = 0.$ Now $g^0(x)=0$ for all $x\in \mathbb{R}$ and $h^0(y)=0$ for all $y\in \mathbb{R}$. So both $g^0$ and $h^0$ are continuous at $0$ and therefore, $f$ is separately continuous at $(0,0)$. However, $f$ is not continuous at $(0,0)$ because $\lim _{(x, y) \rightarrow(0,0)} f(x, y)$ does not exist. We can establish this by finding two sequences $\left(a_j, b_j\right)$ and $\left(\alpha_j, \beta_j\right)$ both of which converge to $(0,0)$ but for which $\lim _{j \rightarrow \infty} f\left(a_j, b_j\right) \neq \lim _{j \rightarrow \infty} f\left(\alpha_j, \beta_j\right) ;$ we also require $\left(a_j, b_j\right) \neq(0,0)$ and $\left(\alpha_j, \beta_j\right) \neq(0,0) \forall j \in \mathbb{N}$. So take, for example, $a_j=\alpha_j=\beta_j=1 / j$ and $b_j=0$. Then $f\left(a_j, b_j\right)=0$ and $f\left(\alpha_j, \beta_j\right)=1 \forall j$ and therefore $\lim _{j \rightarrow \infty} f\left(a_j, b_j\right)=0 \neq 1=\lim _{j \rightarrow \infty} f\left(\alpha_j, \beta_j\right)$. By uniqueness of limits, if $\lim _{(x, y) \rightarrow(0,0)} f(x, y)$ were to exist, $\lim _{j \rightarrow \infty} f\left(a_j, b_j\right)$ and $\lim _{j \rightarrow \infty} f\left(\alpha_j, \beta_j\right)$ would have to have the same value. Since they do not, we conclude that $\lim _{(x, y) \rightarrow(0,0)} f(x, y)$ does not exist.