Does a rational function extend to a point if all its 1-parameter extensions at that point agree? Let $f: X \dashrightarrow Y$ be a rational map where $X$ and $Y$ are varieties over $\mathbb{C}$ with $X$ smooth and $Y$ proper (or projective if necessary). Choose an open set $U \subset X$ on which $f$ is defined.
Pick a closed point $x \in X$ and an algebraic curve $C$ passing through $x$ such that $C \cap U \neq \emptyset$. Since the restriction of $f$ to $C$ can be uniquely extended to all of $C$ we may define $f|_C(x)$.
Let us say that all 1-parameter limits of $f$ at $x$ agree if there exists a point $y \in Y$ such that for any curve $C$ as above $f|_C(x) = y$. 
Could anyone tell me if some variation of the following is true?

If all 1-parameter limits of $f$ agree at $x \in X$ can $f$ be extended to an open set containing $x$?

One suggested alternative in case the first fails:

Does $f$ extend to an open set $\tilde U \subset X$ if for all points in $\tilde U$ all 1-parameter limits of $f$ agree?

 A: Thanks to Jason Starr who answered my question in the comments. I would like to beef up his answer here for future reference.

The answer to the first question--and thus for the second--is in the affirmative. In fact, there are no restrictions on the singularities of $X$; for instance it need not be normal let alone smooth.

Here is the proof:
Step 1
Let $U \subset X$ be any open subset where $f$ is defined, $\Gamma_U \subset U \times Y$ be the graph of $f$ and $\Gamma \subset X\times Y$ be the closure of $\Gamma_U$.
The restricted projection map $\pi: \Gamma \to X$ is proper, moreover it is an isomorphism over $U$. The goal is to show that $\pi$ is in fact an isomorphism over an open set containing $x$. By precomposing with this isomorphism we can extend $f$ around $x$. 
To that end we will show that $\pi$ is quasi-finite of degree 1 around $x$.
Step 2
Let $\Gamma_x$ be the fiber of $\Gamma$ over $x$. Using the valuative criterion of closedness and the universal property of the fiber product it is easy to show that $\Gamma_x$ is in bijection with the set of values that appear as 1-parameter limits of $f$ at $x$. 
The hypothesis states that this latter set consists of a single element. Using semi-continuity we conclude that in a neighbourhood of $x$, the fibers of $\Gamma$ consist of a single element. 
Thus $\pi$ is invertible in a Zariski neighbourhood of $x$ and $f$ extends to this neighbourhood.
