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Section 2.2 Limits and Continuity

For \(1 \le i \le n\text{,}\) the \(i\)-th projection of \(\R^n\) is the function \(\pi_i \colon \R^n \to \R\) defined by
\begin{equation*} \pi_i\langle x_1, \ldots, x_n \rangle = x_i. \end{equation*}
A function \(f \colon A \to \R^m\) from a subset \(A\) of \(\R^n\) is called a function of \(n\) variables. We also call \(f\) a scalar field if \(m=1\) and a vector-valued function if \(m \ge 2\text{.}\) The \(i\)-th component function of \(f\) (\(1 \le i \le m\)) is the composition \(f_i := \pi_i \circ f\text{.}\) Thus
\begin{equation*} f=\langle f_1,\ldots,f_m \rangle. \end{equation*}
In general we use coordinates \(x_1,\ldots,x_n\) on \(\R^n\text{.}\) For \(n \le 3\text{,}\) we will usually write \(x\text{,}\) \(y\text{,}\) and \(z\) instead of \(x_1\text{,}\) \(x_2\text{,}\) and \(x_3\text{.}\)

Definition 2.2.1. Limit.

Let \(A \subseteq \R^n\text{,}\) let \(\va=\langle a_1,\ldots,a_n \rangle\) be an accumulation point of \(A\text{,}\) and let \(f \colon A \to \R^m\text{.}\) For \(L=\langle L_1,\ldots,L_m \rangle \in \R^m\text{,}\) we say that \(f(\vx)\) approaches \(L\) as \(\vx\) approaches \(\va\text{,}\) and write
\begin{equation*} \lim_{\vx \to \va} f(\vx) = L, \end{equation*}
if for every \(\epsilon \gt 0\) there exists \(\delta \gt 0\) such that whenever \(\vx \in A\) and \(0 \lt \norm{\vx-\va} \lt \delta\text{,}\) we have
\begin{equation*} \norm{f(\vx)-L} \lt \epsilon. \end{equation*}
This is the same epsilon-delta idea as in one-variable calculus. The only difference is that both the input and the output may now have several coordinates.

Proof.
If \(\lim_{\vx \to \va} f(\vx)=L\text{,}\) then \(|f_i(\vx)-L_i| \le \norm{f(\vx)-L}\) for each \(i\text{,}\) so every component has the stated limit. Conversely, suppose each \(f_i(\vx)\) tends to \(L_i\text{.}\) Given \(\epsilon \gt 0\text{,}\) choose \(\delta_i \gt 0\) so that \(|f_i(\vx)-L_i| \lt \epsilon/\sqrt{m}\) whenever \(0 \lt \norm{\vx-\va} \lt \delta_i\text{.}\) If \(\delta = \min\{\delta_1,\ldots,\delta_m\}\text{,}\) then
\begin{equation*} \norm{f(\vx)-L}^2 = \sum_{i=1}^m |f_i(\vx)-L_i|^2 \lt \sum_{i=1}^m \frac{\epsilon^2}{m} = \epsilon^2. \end{equation*}
Hence \(\norm{f(\vx)-L} \lt \epsilon\text{.}\)

Example 2.2.3.

Compute
\begin{equation*} \lim_{\langle x,y \rangle\to\langle 1,2 \rangle} \langle x^2-y, xy \rangle. \end{equation*}
Solution.
Let \(f(x,y)=\langle x^2-y, xy \rangle\text{.}\) Its component functions are \(f_1(x,y)=x^2-y\) and \(f_2(x,y)=xy\text{.}\) Since both are polynomials, they are continuous, so
\begin{equation*} \lim_{\langle x,y \rangle\to\langle 1,2 \rangle} f(x,y) = \langle 1^2-2, 1\cdot 2 \rangle = \langle -1,2 \rangle. \end{equation*}

Definition 2.2.4. Continuity.

Let \(f \colon A \to \R^m\) and let \(\va \in A\text{.}\) We say that \(f\) is continuous at \(\va\) if
\begin{equation*} \lim_{\vx \to \va} f(\vx) = f(\va). \end{equation*}
We say that \(f\) is continuous on \(A\) if it is continuous at every point of \(A\text{.}\)
By Proposition 2.2.2, a vector-valued function \(f=\langle f_1,\ldots,f_m \rangle\) is continuous at \(\va\) if and only if each component function \(f_i\) is continuous at \(\va\text{.}\)

Example 2.2.5.

Show that \(r \colon \R \to \R^2\) defined by
\begin{equation*} r(t)=\gv{\cos t, \sin t} \end{equation*}
is continuous.
Solution.
The component functions \(r_1(t)=\cos t\) and \(r_2(t)=\sin t\) are continuous for every \(t \in \R\text{.}\) Hence \(r\) is continuous on \(\R\text{.}\) Its image is the unit circle.
The definition of limit looks formally the same in one variable and in several variables, but the geometry is different. On the real line there are only two ways to approach a point: from the left and from the right. In \(\R^2\) or \(\R^3\text{,}\) there are infinitely many directions and infinitely many curves along which a point can be approached.

Example 2.2.6.

Define \(g \colon \R \setminus \{0\} \to \R\) by
\begin{equation*} g(x) = \begin{cases} -1, & x \lt 0, \\ 1, & x \gt 0. \end{cases} \end{equation*}
Show that \(\lim_{x\to 0} g(x)\) does not exist.
Solution.
As \(x \to 0^-\text{,}\) we have \(g(x)=-1\text{,}\) so the left-hand limit is \(-1\text{.}\) As \(x \to 0^+\text{,}\) we have \(g(x)=1\text{,}\) so the right-hand limit is \(1\text{.}\) Since these two one-sided limits are not equal, \(\lim_{x\to 0} g(x)\) does not exist.
Graph of a jump function equal to negative one for x less than zero and positive one for.
Figure 2.2.7. For a function of one variable, only the left and right approaches matter.

Example 2.2.8.

Define \(h \colon \R^2 \setminus \{\langle 0,0 \rangle\} \to \R\) by
\begin{equation*} h(x,y)=\frac{2xy}{x^2+y^2}. \end{equation*}
Show that \(\lim_{\langle x,y \rangle\to\langle 0,0 \rangle} h(x,y)\) does not exist.
Solution.
If we approach the origin along the line \(y=x\text{,}\) then
\begin{equation*} h(x,x) = \frac{2x^2}{2x^2} = 1 \end{equation*}
for \(x \ne 0\text{.}\) On the other hand, along the line \(y=-x\text{,}\)
\begin{equation*} h(x,-x) = \frac{-2x^2}{2x^2} = -1. \end{equation*}
Since two different paths toward \(\langle 0,0 \rangle\) give two different limiting values, the limit does not exist.
Surface of h, with y equals x at height 1 and y equals negative x at height negative 1.
Figure 2.2.9. A graph of \(h(x,y)=\dfrac{2xy}{x^2+y^2}\text{.}\)
Coordinate axes in the plane with the origin marked..
Figure 2.2.10. In several variables, different paths toward the same point can produce different limiting values.
The previous example shows how to prove that a multivariable limit does not exist: it is enough to find two paths with different limiting values. However, checking a few paths can never prove that a limit exists, because there are infinitely many possible paths.

Example 2.2.11.

Define \(k \colon \R^2 \setminus \{\langle 0,0 \rangle\} \to \R\) by
\begin{equation*} k(x,y)=\frac{x^2y}{x^4+y^2}. \end{equation*}
Show that looking only along straight lines is not enough to determine whether \(\lim_{\langle x,y \rangle\to\langle 0,0 \rangle} k(x,y)\) exists.
Solution.
Along any line \(y=mx\) through the origin,
\begin{equation*} k(x,mx) = \frac{mx^3}{x^4+m^2x^2} = \frac{mx}{x^2+m^2}, \end{equation*}
so \(k(x,mx)\to 0\) as \(x \to 0\text{.}\) If we only tested lines, we might guess the limit is \(0\text{.}\) But along the parabola \(y=x^2\text{,}\)
\begin{equation*} k(x,x^2) = \frac{x^4}{x^4+x^4} = \frac{1}{2} \end{equation*}
for \(x \ne 0\text{.}\) Therefore the limit does not exist.
Surface of k, with y equals 0 at height 0 and y equals x squared at height one half.
Figure 2.2.12. A graph of \(k(x,y)=\dfrac{x^2y}{x^4+y^2}\text{.}\)
In one variable, continuity at a point is controlled by only two nearby behaviors, from the left and from the right. In several variables, a limit must be stable under every nearby direction and every nearby path. This is why continuity in multivariable calculus is subtler, even though the definition looks almost identical.
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