Editing Multivariable calculus (section)

== Limits ==
A study of [[limit of a function|limits]] and [[continuous function|continuity]] in multivariable calculus yields many counterintuitive results not demonstrated by single-variable functions.

A limit along a path may be defined by considering a parametrised path <math>s(t): \mathbb{R} \to \mathbb{R}^n</math> in n-dimensional Euclidean space. Any function <math>f(\overrightarrow{x}): \mathbb{R}^n \to \mathbb{R}^m</math> can then be projected on the path as a 1D function <math>f(s(t))</math>. The limit of <math>f</math> to the point <math>s(t_0)</math> along the path <math>s(t)</math> can hence be defined as

{{NumBlk|:|<math>\lim_{\overrightarrow{x} \to s(t_0)} f(\overrightarrow{x}) = \lim_{t \to t_0} f(s(t))</math>|{{EquationRef|1}}}}

Note that the value of this limit can be dependent on the form of <math>s(t)</math>, i.e. the path chosen, not just the point which the limit approaches.<ref name="CourantJohn1999"/>{{rp|19–22}} For example, consider the function

:<math>f(x,y) = \frac{x^2y}{x^4+y^2}.</math>

If the point <math>(0,0)</math> is approached through the line <math>y=kx</math>, or in parametric form:

[[File:((x^2)(y))⁄((x^4)+(y^2)).png|thumb|Plot of the function {{math|''f''(''x'', ''y'') {{=}} (''x''²y)/(''x''{{sup|4}} + ''y''{{sup|2}})}}]]

{{NumBlk|:|<math>x(t) = t,\, y(t) = kt</math>|{{EquationRef|2}}}}

Then the limit along the path will be:

{{NumBlk|:|<math>\lim_{t \to 0} f(x(t),y(t)) = \lim_{t \to 0} \frac{k t^3}{t^4 + k^2 t^2} = 0</math>|{{EquationRef|3}}}}

On the other hand, if the path <math>y=\pm x^2</math> (or parametrically, <math>x(t)=t,\, y(t)=\pm t^2</math>) is chosen, then the limit becomes:

{{NumBlk|:|<math>\lim_{t \to 0} f(x(t),y(t)) = \lim_{t \to 0} \frac{\pm t^4}{t^4 + t^4} = \pm \frac{1}{2}</math>|{{EquationRef|4}}}}

Since taking different paths towards the same point yields different values, a general limit at the point <math>(0,0)</math> cannot be defined for the function.

A general limit can be defined if the limits to a point along all possible paths converge to the same value, i.e. we say for a function <math>f: \mathbb{R}^n \to \mathbb{R}^m</math> that the limit of <math>f</math> to some point <math>x_0 \in \mathbb{R}^n</math> is L, if and only if

{{NumBlk|:|<math>\lim_{t \to t_0} f(s(t)) = L</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>.

=== 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.

===Theorems regarding multivariate limits and continuity ===
* All properties of linearity and superposition from single-variable calculus carry over to multivariate calculus.
* '''Composition''': If <math>f: \mathbb{R}^n \to \mathbb{R}^m</math> and <math>g: \mathbb{R}^m \to \mathbb{R}^p</math> are both multivariate continuous functions at the points <math>x_0 \in \mathbb{R}^n</math> and <math>f(x_0) \in \mathbb{R}^m</math> respectively, then <math>g \circ f: \mathbb{R}^n \to \mathbb{R}^p</math> is also a multivariate continuous function at the point <math>x_0</math>.
* '''Multiplication''': If <math>f: \mathbb{R}^n \to \mathbb{R}</math> and <math>g: \mathbb{R}^n \to \mathbb{R}</math> are both continuous functions at the point <math>x_0 \in \mathbb{R}^n</math>, then <math>fg: \mathbb{R}^n \to \mathbb{R}</math> is continuous at <math>x_0</math>, and <math>f/g : \mathbb{R}^n \to \mathbb{R}</math> is also continuous at <math>x_0</math> provided that <math>g(x_0) \neq 0</math>.
* If <math>f: \mathbb{R}^n \to \mathbb{R}</math> is a continuous function at point <math>x_0 \in \mathbb{R}^n</math>, then <math>|f|</math> is also continuous at the same point.
* If <math>f: \mathbb{R}^n \to \mathbb{R}^m</math> is [[Lipschitz continuous]] (with the appropriate normed spaces as needed) in the neighbourhood of the point <math>x_0 \in \mathbb{R}^n</math>, then <math>f</math> is multivariate continuous at <math>x_0</math>.

{{Collapse top|Proof|expand=true}}
From the Lipschitz continuity condition for <math>f</math> we have

{{NumBlk|:|<math>|f(s(t))-f(s(t_0))| \leq K|s(t)-s(t_0)|</math>|{{EquationRef|8}}}}

where <math>K</math> is the Lipschitz constant. Note also that, as <math>s(t)</math> is continuous at <math>t_0</math>, for every <math>\delta > 0</math> there exists a <math>\epsilon > 0</math> such that <math>|s(t)-s(t_0)| < \delta</math> <math>\forall |t-t_0| < \epsilon</math>.

Hence, for every <math>\alpha > 0</math>, choose <math>\delta = \frac{\alpha}{K}</math>; there exists an <math>\epsilon > 0</math> such that for all <math>t</math> satisfying <math>|t-t_0| < \epsilon</math>, <math>|s(t)-s(t_0)| < \delta</math>, and <math>|f(s(t)) - f(s(t_0))| \leq K|s(t)-s(t_0)| < K\delta = \alpha</math>. Hence <math>\lim_{t \to t_0} f(s(t))</math> converges to <math>f(s(t_0))</math> regardless of the precise form of <math>s(t)</math>.

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