The Squeeze Theorem

Section 12.2 The Squeeze Theorem

It is clear from the graph below that \(\limit{x}{\infty}{\sin(x)}\) does not exist since the \(\sin(x)\) has no horizontal asymptotes. To say that this limit does not exist means that \(y=\sin(x)\) is not approaching any particular value as \(x\rightarrow\infty\text{.}\) It’s not even “approaching” \(+\infty\) or \(-\infty\) since it is bounded both above and below.

Now how about this limit:

\begin{equation*} \tlimit{x}{\infty}{\frac{\sin(x)}{x}}? \end{equation*}

Since\(\limit{x}{\infty}{\sin(x)}\) does not exist we can’t apply Theorem 12.1.8. Neither is it of the form \(\frac{\approach{\infty}}{\approach{\infty}},\) so none of the techniques we’ve used before will help us.

We need a new idea.

Look at the graph of \(f(x) = \frac{\sin(x)}{x}\text{:}\)

Can you see that the horizontal asymptote is \(y=0\text{?}\)

It certainly looks like \(y=0\) is an asymptote in both directions which seems to suggest that

\begin{equation*} \tlimit{x}{\infty}{\frac{\sin(x)}{x}} = \tlimit{x}{-\infty}{\frac{\sin(x)}{x}} = 0\text{.} \end{equation*}

This is correct, but is there a way to see this algebraically? After all, this graph could just as easily support the conclusion that \(y=0.00000001\) (or \(y=\) any other number which is extremely close to zero) is the asymptote. We don’t want to draw conclusions from a picture, but it is difficult to see what we can do, isn’t it?

We will analyze what happens when \(x\rightarrow\infty\) and leave the case \(x\rightarrow-\infty\) for you in Problem 12.2.1.

Let’s “disassemble” our function in order to understand it better. The function \(f(x)=\frac{\sin(x)}{x}\) clearly comes in two parts: \(\sin(x)\) and \(\frac{1}{x}\text{.}\) Because

\begin{equation*} -1\le\sin(x)\le1 \end{equation*}

we know that

\begin{equation*} -\frac{1}{x}\le\frac{\sin(x)}{x}\le\frac1x. \end{equation*}

If we graph \(y=\frac{\sin(x)}{x}\text{,}\) \(y=\frac1x\text{,}\) and \(y=-\frac1x\) together as in the sketch below we see that for positive values of \(x\) the graph of \(y=\frac{\sin(x)}{x}\) is always caught between the graph of \(y=\frac1x\) and \(y=-\frac1x\text{,}\) as seen below.

We know that both \(\pm\frac1x\rightarrow0\) as \(x\rightarrow\infty\text{.}\) Since \(\frac{\sin(x)}{x}\) is caught between \(\frac1x\) and \(-\frac1x\) it follows that, for positive values of \(x\text{,}\) the highest and lowest points on the graph of \(\frac{\sin(x)}{x}\) must also approach zero as \(x \rightarrow\infty\text{.}\) Therefore \(\tlimit{x}{\infty}{\frac{\sin(x)}{x}}=0\text{.}\)

Problem 12.2.1.

Show that for negative values of \(x\text{,}\) \(-\frac1x \le \frac{\sin(x)}{x} \le \frac1x\) also, so that \(\tlimit{x}{-\infty}{\frac{\sin(x)}{x}=0}\) as well.

This idea is often called the Squeeze Theorem or the Sandwich Theorem.

Theorem 12.2.2. The Squeeze Theorem at Infinity.

There are two cases:

If \(\alpha(x)\le f(x)\le \beta(x)\) on some interval, \((c, \infty)\) and

\begin{equation*} \limit{x}{\infty}{\alpha(x)} = L = \limit{x}{\infty}{\beta(x)} \end{equation*}

then \(\limit{x}{\infty}{f(x)} = L\) also.
If \(\alpha(x)\le f(x)\le \beta(x)\) on some interval, \((-\infty,c)\) and

\begin{equation*} \limit{x}{-\infty}{\alpha(x)} = L = \limit{x}{-\infty}{\beta(x)} \end{equation*}

then \(\limit{x}{-\infty}{f(x)} = L\) also.

In the previous example, notice that we had \(\frac{-1}{x}\le \frac{\sin(x)}{x}\le\frac1x\) for all \(x\neq0\text{,}\) but this is more than is actually required. We only need for the inequality to be satisfied when \(x\) sufficiently large, in either the positive or negative direction. That’s why we only require that \(\alpha(x)\le f(x)\le \beta(x)\) on the interval \((c,\infty)\) (positive case) or \((-\infty,c)\) (negative case) rather than for all values of \(x\text{.}\)

Problem 12.2.3.

Use the Squeeze Theorem to determine each of the following limits:

(a)

\(\limit{x}{\infty}{\frac{x\cos(x)}{x^2+1}}\)

(b)

\(\limit{x}{-\infty}{\frac{x\cos(x)}{x^2+1}}\)

(c)

\(\limit{x}{\infty}{\frac{\lfloor x\rfloor}{x^2}},\) where \(\lfloor x\rfloor\) represents the greatest integer less than or equal to \(x\text{.}\)

Back in Section 8.13 you showed that \(y(t)=e^{-\frac{t}{2}}\left[\cos\left(\frac{\sqrt{3}}{2}t\right)+\sin\left(\frac{\sqrt{3}}{2}t\right)\right]\) satisfies the differential equation which models damped oscillation (oscillation with resistance):

\begin{equation*} \dfdxn{y}{t}{2}+\dfdx{y}{t}+y=0. \end{equation*}

Since \(y(t)\) is called a damped oscillator what do you expect \(\limit{t}{\infty}{y(t)}\) to be equal to? Guess, if you aren’t sure.

Drill 12.2.4.

Use the squeeze theorem to compute

\begin{equation*} \limit{t}{\infty}{e^{-\frac{t}{2}}\left[\cos\left(\frac{\sqrt{3}}{2}t\right)+\sin\left(\frac{\sqrt{3}}{2}t\right)\right]}. \end{equation*}

Is this consistent with your guess? Explain.

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