Editing Multivariable calculus (section)

=== Continuity ===
From the concept of limit along a path, we can then derive the definition for multivariate continuity in the same manner, that is: we say for a function <math>f: \mathbb{R}^n \to \mathbb{R}^m</math> that <math>f</math> is continuous at the point <math>x_0</math>, if and only if

{{NumBlk|:|<math>\lim_{t \to t_0} f(s(t)) = f(x_0)</math>|{{EquationRef|5}}}}

for all continuous functions <math>s(t): \mathbb{R} \to \mathbb{R}^n</math> such that <math>s(t_0)=x_0</math>.

As with limits, being continuous along ''one'' path <math>s(t)</math> does not imply multivariate continuity.

Continuity in each argument not being sufficient for multivariate continuity can also be seen from the following example.<ref name="CourantJohn1999" />{{rp|17–19}} For example, for a real-valued function <math>f: \mathbb{R}^2 \to \mathbb{R}</math> with two real-valued parameters, <math>f(x,y)</math>, continuity of <math>f</math> in <math>x</math> for fixed <math>y</math> and continuity of <math>f</math> in <math>y</math> for fixed <math>x</math> does not imply continuity of <math>f</math>.

Consider
:<math>
f(x,y)=
\begin{cases}
\frac{y}{x}-y & \text{if}\quad 0 \leq y < x \leq 1 \\
\frac{x}{y}-x & \text{if}\quad 0 \leq x < y \leq 1 \\
1-x & \text{if}\quad 0 < x=y \\
0 & \text{everywhere else}.
\end{cases}
</math>

It is easy to verify that this function is zero by definition on the boundary and outside of the quadrangle <math>(0,1)\times (0,1)</math>. Furthermore, the functions defined for constant <math>x</math> and <math>y</math> and <math>0 \le a \le 1</math> by
:<math>g_a(x) = f(x,a)\quad</math> and <math>\quad h_a(y) = f(a,y)\quad</math>
are continuous. Specifically,
:<math>g_0(x) = f(x,0) = h_0(0,y) = f(0,y) = 0</math> for all {{mvar|x}} and {{mvar|y}}. Therefore, <math>f(0,0)=0</math> and moreover, along the coordinate axes, <math>\lim_{x \to 0} f(x,0) = 0</math> and <math>\lim_{y \to 0} f(0,y) = 0</math>. Therefore the function is continuous along both individual arguments.

However, consider the parametric path <math>x(t) = t,\, y(t) = t</math>. The parametric function becomes

{{NumBlk|:|<math>
f(x(t),y(t))=
\begin{cases}
1-t & \text{if}\quad t > 0 \\
0 & \text{everywhere else}.
\end{cases}
</math>|{{EquationRef|6}}}}

Therefore,

{{NumBlk|:|<math>\lim_{t \to 0^+} f(x(t),y(t)) = 1 \neq f(0,0) = 0</math>|{{EquationRef|7}}}}

It is hence clear that the function is not multivariate continuous, despite being continuous in both coordinates.