Section 1.8 The Squeeze Theorem
Objectives
- Evaluate the limit of a function by using the squeeze theorem.
- Articulate the difference between a one-sided and two-sided limit.
- Evaluate two-sided limits using two one-sided limits.
lim
we recognize that
\begin{equation*}
\frac{(x-2)(x-3)}{x-2}=x-3
\end{equation*}
at all values of x except possibly at x=2\text{.} So the two functions have the same limit there, and we can use limit laws to evaluate
\begin{equation*}
\lim_{x \rightarrow 2} x-3 = 2-3=-1.
\end{equation*}
The squeeze theorem combines this idea with approximation. Suppose we have a limit we want to find,
\begin{equation*}
\lim_{x \rightarrow 0}x \cos(x).
\end{equation*}
We want to find two functions for which we can compute the limit, call them f(x) and h(x) that bracket our function. In other words we want
\begin{equation*}
f(x) \le x \cos(x) \le h(x)
\end{equation*}
for all values nearby to zero. If \lim_{x \rightarrow 0}f(x)=L and \lim_{x \rightarrow 0} h(x)=M and we take the limit of all the parts of this set of inequalities, then we get
\begin{equation*}
L=\lim_{x \rightarrow 0}f(x) \le \lim_{x \rightarrow 0}x \cos(x) \le \lim_{x \rightarrow 0}h(x)=M,
\end{equation*}
or
\begin{equation*}
L \le \lim_{x \rightarrow 0}x \cos(x) \le M.
\end{equation*}
Now this tells us something about the limit - it is between L and M - but still doesn't give us the complete answer. The trick now is to find an f(x) and g(x) so that L=M - then the computation above would imply that \lim_{x \rightarrow 0}x \cos(x)=L\text{.}
But how in the world can we find these functions? Sadly, there is no formula or recipe for doing this - we need to think about x and \cos(x) and see what we can figure out. Let's start by looking at the graph of \cos(x) over one of its periods in Figure 1.8.1. 
\begin{equation*}
-1\cdot x \le x \cos(x) \le 1 \cdot x,
\end{equation*}
and for negative values of x, we have
\begin{equation*}
1 \cdot x \le x \cos(x) \le -1 \cdot x.
\end{equation*}
This makes this a little complicated - to help us we will introduce a new notion, the one sided limit. When we made our somewhat informal definition of the limit, we thought about what value the function tended to as we approached the x value from both sides. A one-sided limit has a similar definition but we only consider values from the right or left of the limiting x value. We use a modified notation as well:
\begin{equation*}
\lim_{x \rightarrow a^+}f(x)
\end{equation*}
when we only look at values to the right of a and
\begin{equation*}
\lim_{x \rightarrow a^-}f(x)
\end{equation*}
when we only consider values to the left. The two-sided limit exists if both one-sided limits exist and are equal to the same value.
Let's apply this idea here to try to find \lim_{x \rightarrow 0}x \cos(x)\text{.} First, we'll look at values to the right of zero - positive values of x\text{.} Taking the one-sided limit of the series of inequalities, we have
\begin{equation*}
\lim_{x \rightarrow 0^+}-x \le \lim_{x \rightarrow 0^+}x \cos(x) \le \lim_{x \rightarrow 0^+}x.
\end{equation*}
Now both the functions f(x)=-x and h(x)=x are polynomials so we can find their limits by evaluation:
\begin{equation*}
\lim_{x \rightarrow 0^+}-x=0=\lim_{x \rightarrow 0^+}x.
\end{equation*}
Since these are equal, we conclude
\begin{equation*}
\lim_{x \rightarrow 0^+}x\cos(x)=0.
\end{equation*}
If we only consider values to the left- negative x values- we have a similar computation:
\begin{equation*}
0=\lim_{x \rightarrow 0^-}-x \le \lim_{x \rightarrow 0^-}x \cos(x) \le \lim_{x \rightarrow 0^-}x=0,
\end{equation*}
and we conclude that
\begin{equation*}
\lim_{x \rightarrow 0^-}x \cos(x)=0
\end{equation*}
as well. Taken together, this tells us that the limit of x \cos(x) as x approaches zero is zero.
This is the essence of the Squeeze Theorem: Theorem 1.8.2. The Squeeze Theorem.
Let f(x),g(x)\text{,} and h(x) be defined for all x \neq a over an open interval containing a\text{.} If
\begin{equation*}
f(x)\le g(x)\le h(x)
\end{equation*}
for all x \neq a in an open interval containing a and
\begin{equation*}
\lim_{x \rightarrow a} f(x)=L=\lim_{x \rightarrow a} h(x),
\end{equation*}
where L is a real number, then \lim_{x \rightarrow a} f(x)=L\text{.}