\documentclass{amsart}

\newtheorem*{theorem}{Theorem}
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\newcommand{\N}{\mathbf{N}}

\title{Substitutions in Limits}

\author{Math 166: Calculus II}

\date{Spring 2000}

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\maketitle

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In this note we examine how to make a ``change of variables'' or a
``substitution'' when calculating limits.  In the following theorem,
we assume that the substitution function $u = g(x)$ is continuous and
non-constant, but we do not to assume anything about the function it is
composed with.

\begin{theorem}
Let $g(x)$ be a function defined on an interval $[a,b]$. Assume that
$g(x)$ is continuous and non-constant on any subinterval of $[a,b]$
containing $p$.  Define $p^*$ to be the symbol $p^+$ if $p=a$, or $p^-$
if $p=b$, or $p$ otherwise. Similarly, let $q=g(p)$ and define $q^*$
to be the symbol $q^+$ if $g(x)$ has a relative minimum at $p$, or $q^-$
if $g(x)$ is a relative maximum at $p$, or $q$ otherwise.  Then for any
function $f(u)$ defined in an appropriate neighborhood of $q$ we have
\[ \lim_{x\to p*}f(g(x)) = \lim_{u\to q^*} f(u) \]
\end{theorem}

\begin{proof}
We must show the existence of one limit implies the existence of the
other and that the two limiting values are equal.

Let us first assume $\lim_{x\to p^*}f(g(x))=L$ exists and prove
$\lim_{u\to q^*} f(u)=L$.  This part is similar to the proof of
Theorem~3.10, Apostol p.147.  By the definition of limit, given
$\epsilon>0$, there exists an interval $I$ containing $p$ such that if
$x\in I$, then $|f(g(x))-L|<\epsilon$. If $g(x)$ has a relative extremum
at $x=p$, we replace $I$ by a smaller interval containing $p$ so that so
that $q=g(p)$ is an extreme value of $g(x)$ restricted to~$I$.  Let $J$
be a closed subinterval of $I$ containing $p$.  Since $g(x)$ is continuous
and non-constant on any subinterval containing $p$, the Extreme Value
Theorem and the Intermediate Value Theorem imply that the range of $g(x)$
restricted to $J$ is also a closed interval, say $K=[c,d]$. Note that
if $g(x)$ has a relative minimum at $p$ then $q=c$, or if $g(x)$ has a
relative maximum at $p$ then $q=d$.  By construction, for any $u\in K$,
there is an $x\in J\subset I$ such that $u=g(x)$.  Then, $|f(u) - L|
= |f(g(x)) - L| < \epsilon$, since $x\in I$. This proves $\lim_{u\to
q^*} f(u)=L$.

Let us now assume $\lim_{u\to q^*} f(u)=L$ exists and prove $\lim_{x\to
p^*}f(g(x))=L$. This part is similar to the proof of Theorem~3.5,
Apostol p.141.  By the definition of limit, given $\epsilon>0$, there
exists an interval $K$ containing $q$ such that if $u\in K$ then
$|f(u)-L|<\epsilon$.  Since $g(x)$ is continuous at $p$ (and this is
all we need to assume about $g(x)$ for this part of the proof), there
is an interval $I$ containing $p$ such that if $x\in I$ then $g(x)\in
K$, which implies $|f(g(x)) - L|<\epsilon$.  This proves $\lim_{x\to
p^*}f(g(x))=L$.
\end{proof}

\subsection*{Example} $\lim_{x\to 0} \frac{\sin(x^2)}{x^2} = \lim_{u\to
0^+} \frac{\sin(u)}{u} = 1$

\subsection*{Example} To see why the assumptions of the theorem are
necessary, consider the function
\[
g(x) = \left\{\begin{array}{ll}
\frac{1}{n} & \mbox{if $\frac{1}{n+1}<|x| \le \frac{1}{n}$ for some $n\in\N$}\\
\\
0           & \mbox{otherwise}
\end{array}\right.
\]
Then $g(x)$ is non-constant in any interval containing $0$ and is
also continuous at~$0$ since $\lim_{x\to 0}g(x) = g(0) = 0$.  However,
$g(x)$ is not continuous in any \emph{interval} containing~$0$ since
for $n\in\N$,
\[
\lim_{x\to\frac{1}{n}^-} g(x) = \frac{1}{n}
\ne \frac{1}{n-1} = \lim_{x\to\frac{1}{n}^+} g(x)
\]
If we define
\[
f(u) = \left\{\begin{array}{ll}
1 & \mbox{if $|u| = \frac{1}{n}$ for some $n\in\N$}\\
\\
0 & \mbox{otherwise}
\end{array}\right.
\]
then clearly $f(g(x)) = 1$ for all $0<|x|\le 1$ so $\lim_{x\to 0}
f(g(x)) = 1$. Yet $\lim_{u\to 0^+}f(u)$ does not exist since $f(u)$
takes on the values 0 and 1 for arbitrarily small positive values of $u$.

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