Problem Statement: Let $U\subset \mathbb{R}^{m}$ be open, and $f:U\rightarrow\mathbb{R}^{n}$. Assume that for any continuous curve $\gamma:[0,1]\rightarrow U$ with $\gamma(0)=x_{0}\in U$, we have $\lim_{t\rightarrow 0^{+}} f(\gamma(t))=L\in \mathbb{R}^{n}$. Then $\lim_{x\rightarrow x_{0}} f(x_{k})=L$.
I am not quite sure how to approach this proof. I thought that if I can show that $f$ is continuous in $U$, then that would be enough to finish the proof, but I feel that we do not have enough information about $f$ to show it is continuous in $U$.
My professor gave us a hint that it is enough to show that for any sequence $\left\{x_{k}\right\}\in U$ with $\lim x_{k}=x_{0}$, then $f(x_{k})\rightarrow L$.
I think it makes sense to me why this should be true, but the only thing that I feel is the missing piece is that we don't know if $f$ is continuous in $U$.
Is this the correct strategy? Any suggestions on how to approach this problem are appreciated!
You don't need to show that $f$ is continuous in all of $U$ (which is a lot to ask of $f$), but your professor's hint can be quite useful.
You want to show that for any sequence $\{x_k\}$ of points in $U$ with $\lim_{k\to\infty} x_k=x_0$ we have $\lim_{k\to\infty} f(x_k)= L$. It would be nice if we had a continuous path $\gamma\colon[0,1]\to U$ so that $\gamma(0)=x_0$ and so that we could choose a sequence of values $\{t_k\}$ which decrease to $0$ so that $\gamma(t_k)=x_k$. Then we could use the fact that \begin{equation} \lim_{t\to 0^+} f(\gamma(t)) = L \end{equation} to show that $\lim_{k\to\infty} f(x_k)=L$. Consider letting \begin{equation} t_k = 2^{1-k}, \end{equation} setting $\gamma(t_k)=x_k$, and letting $\gamma$ follow any continuous path in $U$ from $x_k$ to $x_{k+1}$ as $t$ varies from $t_k$ to $t_{k+1}$. (Edit: For sufficiently large $K$, all the points $x_k$ with $k\geq K$ will be contained in some convex neighborhood $V\subset U$, so that the path from $x_k$ to $x_{k+1}$ can be chosen to be a line segment. This will avoid the problems mentioned by zhw in the comments.) This will define $\gamma$ on $(0,1]$, and then you'll need to show that if we set $\gamma(0)=x_0$, $\gamma$ is indeed a continuous path.