characterisation of continuity in $\mathbb{R}^n$ I want to know if the following condition is sufficient condition for continuity in higher dimension.
Let's take 2 dimension, for example, let's take a continuous function
Let $\phi(t)$ be a continuous curve $\phi(t) = (x_t,y_t)$ such that $\phi(0)=(x^*,y^*)$.
If $f(\phi(t)) \rightarrow f(x^*,y^*)$ as $t\rightarrow 0$ for any choice of $\phi(t)$, does this imply $f$ is continuous at $(x^*,y^*)$?
 A: Yes, this is true.  You can prove this by interpolating sequences with curves.  To show $f$ is continuous at $(x^*,y^*)$ you need to prove that if $(x_n,y_n)\to (x^*,y^*)$, then $f(x_n,y_n)\to (x^*,y^*)$.  To prove this, you can choose a continuous curve $\phi:[0,1]\to \mathbb{R}^2$ such that $\phi(0)=(x^*,y^*)$ and $\phi(1/n)=(x_n,y_n)$ for all positive integers $n$.  Then $f(\phi(t))\to f(x^*,y^*)$ as $t\to 0$ implies that $f(\phi(1/n))\to f^(x^*,y^*)$, which is exactly what we want.
It remains to show that such a $\phi$ actually exists.  We can construct such a $\phi$ by just interpolating linearly: for $t\in [1/(n+1),1/n]$, let $\phi(t)$ move linearly from $(x_{n+1},y_{n+1})$ to $(x_n,y_n)$.  To check continuity of $\phi$ at $0$, just note that for any $\epsilon>0$ there exists $N$ such that $(x_n,y_n)$ is within $\epsilon$ of $(x^*,y^*)$ for all $n\geq N$, which means $\phi(t)$ is within $\epsilon$ of $(x^*,y^*)$ for all $t\leq 1/n$ since the ball of radius $\epsilon$ around $(x^*,y^*)$ is convex.
The same argument works in $\mathbb{R}^n$ for any $n$.
