#math/calculus
A multivariable limit is **one joint limit**, not one limit taken separately in each coordinate. Let
$
p=(x_0,y_0),
\qquad
h=(\Delta x,\Delta y)=(x-x_0,y-y_0).
$
Then
$
(x,y)\to p
\quad\Longleftrightarrow\quad
\|h\|=\sqrt{(\Delta x)^2+(\Delta y)^2}\to0.
$
Thus $\Delta x$ and $\Delta y$ become small together, with no prescribed order or relationship between them. This includes approaches along either coordinate axis, every line, and every curved path.
Using Euclidean distance,
$
\lim_{(x,y)\to(x_0,y_0)}f(x,y)=L
$
means
$
\forall\varepsilon>0\ \exists\delta>0:
0<\|(x,y)-(x_0,y_0)\|<\delta
\Longrightarrow |f(x,y)-L|<\varepsilon.
$
![[tikz-e2671c0e703b.svg]]
%% tikz-source
```latex
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}[>=Stealth,scale=1.0,line cap=round]
\draw[->,black!45] (-.5,0)--(5.4,0) node[right]{$x$};
\draw[->,black!45] (0,-.5)--(0,4.4) node[above]{$y$};
\coordinate (P) at (3.0,2.0);
\filldraw[fill=blue!10,draw=blue!70!black,very thick] (P) circle (1.12);
\fill[white] (P) circle (3pt);
\draw[blue!70!black,very thick] (P) circle (3pt);
\draw[->,red!75!black,thick] (P)--++(.79,.79) node[midway,below right]{$\delta$};
\node[below left] at (P) {$(x_0,y_0)$};
\node[align=left] at (7.2,2.0)
{$0<\lVert(x,y)-(x_0,y_0)\rVert<\delta$\\forces $|f(x,y)-L|<\varepsilon$};
\end{tikzpicture}
\end{document}
```
%%
*The punctured $\delta$-disk contains every way of approaching $(x_0,y_0)$ in the input plane.*
Taking $\Delta x\to0$ with $\Delta y=0$ checks only the $x$-axis; taking $\Delta y\to0$ with $\Delta x=0$ checks only the $y$-axis. Even if these two limits agree, the joint limit may fail. For example,
$
f(x,y)=\frac{xy}{x^2+y^2}
$
equals $0$ on both axes, but equals $\tfrac12$ along $y=x$. Therefore its joint limit at the origin does not exist.
In general, two paths with different limiting values disprove a joint limit. Agreement along any finite collection of paths does not prove one; the $\varepsilon$-$\delta$ condition must control all sufficiently small displacements at once.
$f$ is `continuous` at $(x_0,y_0)$ when
$
\lim_{(x,y)\to(x_0,y_0)}f(x,y)=f(x_0,y_0).
$
Sums, products, compositions, and nonzero-denominator quotients of continuous functions are continuous.
## Connection to differentiability
In [[15B1 - Linear Approximation and Differentiability]], define the error in the proposed linear approximation by
$
R(h)=f(p+h)-f(p)-Df(p)h.
$
The approximation is valid to first order precisely when the **joint** limit
$
\lim_{h\to0}\frac{R(h)}{\|h\|}=0
$
holds. Computing $f_x(p)$ and $f_y(p)$ checks coordinate-direction slopes; it does not by itself control $R(h)$ along all approaches. Continuous first partial derivatives near $p$ are a convenient sufficient condition that does provide this control.