Limits and Continuity

1. General Metric Space Formulation

When \((X,d_X)\) and \((Y,d_Y)\) are general topological spaces and \(f : X \rightarrow Y\),

\[ \lim_{x \rightarrow p} f(x) = y \]

when for all \(\epsilon > 0\), there exists \(\delta > 0\) such that every \(x \in B_\delta(p) \setminus \{p\}\) satisfies \(d_Y(f(x),y) < \epsilon\). We say \(f\) is continuous at \(p\) when \(\lim_{x \rightarrow p} f(x) = f(p)\).

Theorem (Uniqueness of Limits)

If \(p\) is a limit point of \(X\) (meaning that every neighborhood of \(p\) in \(X\) contains a point other than \(p\) itself), then

\[ \lim_{x \rightarrow p} f(x) = y \text{ and } \lim_{x \rightarrow p} f(x) = y' \]

implies \(y = y'\).

Proof

Use the existence of both limits to show that for every \(\epsilon > 0\) there is some point \(x \neq p\) such that \(d(x,y') < \epsilon/2\) and \(d(x,y) < \epsilon/2\). This means by the triangle inequality that \(d(y,y') < \epsilon\) for all positive \(\epsilon\).

Theorem (Composition)

If \(f : X \rightarrow Y\) and \(g : Y \rightarrow Z\) for metric spaces \((X,d_X), (Y,d_Y),\) and \((Z,d_Z)\) and if

\[ \lim_{x \rightarrow p} f(x) = q \text{ and } \lim_{y \rightarrow q} g(y) = g(q) = r \]

then

\[ \lim_{x \rightarrow p} g(f(x)) = r. \]

In particular, the limit must exist.

Proof

Fix \(\epsilon\); let \(\eta\) be such that \(d_Y(y,q) < \eta\) implies \(d_Z(g(q),g(y)) < \epsilon\). Now for this \(\eta\), let \(\delta\) be such that \(x \in B_\delta(p) \setminus \{p\}\) implies \(d_Y(f(x),q) < \eta\). Chain the implications together to complete.

Theorem (Sequential Characterization)

If \((X,d_X)\) and \((Y,d_Y)\) are metric spaces and \(f : X \rightarrow Y\), then

\[ \lim_{x \rightarrow p} f(x) \]

exists and equals \(y\) if and only if \(\lim_{n \rightarrow \infty} f(x_n) = y\) for all sequences in \(X \setminus \{p\}\) which converge to \(p\).

Proof

The forward direction is the simpler one to prove. For the reverse direction, suppose that \(\lim_{x \rightarrow p} f(x)\) does not exist. Then there must be some \(\epsilon > 0\) such that every \(\delta > 0\) admits some point \(x_\delta\) in \(B_{\delta}(p) \setminus \{p\}\) such that \(d_Y(f(x_\delta),y) \geq \epsilon\). Consider the sequence of \(x\)'s obtained by taking \(\delta = 1,\frac{1}{2}, \frac{1}{3}, \frac{1}{4},\ldots\).

2. Limits in \({\mathbb R}^n\)

Theorem

Suppose \(E \subset {\mathbb R}^n\) and \(f : E \rightarrow {\mathbb R}^m\). Let \(f\) be written in standard coordinates as \(f(x) := (f_1(x),\ldots,f_m(x))\) for \(f_1,\ldots,f_m\) being real-valued functions on \(E\). Then for any limit point \(p\) of \(E\),

\[ \lim_{x \rightarrow p} f(x) = (\lim_{x \rightarrow p} f_1(x),\ldots,\lim_{x \rightarrow p} f_m(x)). \]

In particular, the limit on the left exists if and only if each limit on the right exists.

Important

The theorem above shows that the influence of the dimension \(m\) of the range space is rather trivial–one can simply treat the function \(f\) as an \(m\)-tuple of real-valued functions. There is no analogous way to “trivialize” the dependence of \(f\) on the dimension \(n\) of the domain. These are the examples like

\[ \lim_{(x,y) \rightarrow (0,0)} \frac{xy}{x^2+y^2}\]

frequently encountered in multivariable calculus classes, which show that limits existing along lines, parabolas, etc. are not sufficient to guarantee the existence of the full limit.

Theorem (Continuity of Algebraic Operations)

The functions \(s(x,y) := x+y\) and \(p(x,y) := xy\) are continuous from \({\mathbb R}^2\) to \({\mathbb R}\) at all points. The function \(r(x,y) := x/y\) is continuous on \({\mathbb R} \times ({\mathbb R} \setminus \{0\})\).

Corollary

For real-valued functions \(f,g\) on a metric space \((X,d)\) if \(\lim_{x \rightarrow p} f(x) = a\) and \(\lim_{x \rightarrow p} g(x) = b\), then

\(\displaystyle \lim_{x \rightarrow p} (f(x) + g(x)) = a + b\)
\(\displaystyle \lim_{x \rightarrow p} f(x) g(x) = ab\)
\(\displaystyle \lim_{x \rightarrow p} \frac{f(x)}{g(x)} = \frac{a}{b}\) if \(b \neq 0\) In particular, each limit must exist.

Proof

Note that for the last inequality, you begin by demonstrating that continuity of \(g\) at \(p\) combined with \(g(p) \neq 0\) mean that \(g\) is necessarily nonzero on some \(B_\delta(p) \setminus \{p\}\) so that the limit makes sense.