Let’s look briefly at the **Squeeze Theorem**, or the **Sandwich Theorem**, that lets us prove limit results for certain functions. We won’t use this approach often, because there aren’t very many routine problems that you can solve with it — but it should be in your toolkit.

The “Squeeze” or “Sandwich” names are apt, because the theorem says that if your function always lies between two other functions near the point of interest, and those functions have equal limits there, then your function must have the same limit because it’s “squeezed” between the other two.

If $g(x) \le f(x) \le h(x)$ when

\[\lim_{x \to a}g(x) = \lim_{x \to a}h(x) = L\] then

\[\lim_{x \to a}f(x) = L\]

The following example illustrates. We’re going to revisit a limit we looked at earlier, when we used Desmos graphs to examine “Some Limits That Do Exist; Some That Do Not”: $\displaystyle{ \lim_{x \to 0}x \cos\left(\dfrac{1}{x} \right)}.$ You might recall the discussion there between Luis and Qiao about whether the limit exists and equals 0, or does not exist at all because $\cos(1/x)$ oscillates wildly near $x=0.$. Let’s now reason about the limit without relying on Desmos.

Example 1: $\displaystyle{ \lim_{x \to 0}x \cos(1/x)}$

Prove $\displaystyle{ \lim_{x \to 0}x \cos(1/x)= 0}.$

*Solution*.

Looking at the function, you probably recall that $\displaystyle{ \lim_{x \to 0}\cos(1/x)}$ does not exist because the cos oscillates wildly near $x=0,$ so there is no single number you can get as close to as you’d like as $x \to 0.$ On the other hand, $\displaystyle{ \lim_{x \to 0}x =0}.$ So what is the limit of the product of one function whose limit doesn’t exist, and another whose limit is 0?

The answer is: it depends. You can’t immediately draw a conclusion in this situation, and have to reason more fully.

In this case, the key realization is that the cosine function *always* oscillates between $-1$ and 1, and never exceeds those bounds:

\[-1 \le \cos(1/x) \le 1 \]
Hence when we multiply every term in the inequality by *x,* we have

\[-x \le x\cos(1/x) \le x\]
We’ll look graphically at the preceding line in a minute. For now, notice that we now have the requirement of the Squeeze Theorem in place: we know our function of interest lies between two functions whose limiting behavior at $x=0$ we know about:

\[\lim_{x \to 0}(-x) = 0 \quad \text{and} \quad \lim_{x \to 0}x =0 \]
Hence by the Squeeze Theorem,

\[\lim_{x \to 0}x\cos(1/x) = 0 \quad \cmark\]
The interactive Desmos calculator shows the function $f(x) = x\cos(1/x)$ as a solid red curve, while the green dashed lines indicate $y=x$ and $y=-x.$ You can see how near $x=0$ the *x* acts as an “envelope” for the cosine function, bounding (or “squeezing”) its amplitude. Using the buttons beneath the graph, you can view the original function $f(x),$ or just $\cos(1/x),$ or just the lines $y = \pm x.$

Graph of $f(x) = x\cos(1/x)$ to explore limit near $x=0$

Note in particular how $\displaystyle{\lim_{x \to 0}\cos(1/x)}$ does not exist, but the “squeezed” function $f(x) = x\cos(1/x)$ is forced to 0 as $x \to 0.$

Because it’s of such limited use, problems requiring the Squeeze Theorem don’t appear frequently on exams and sometimes isn’t even required for any homework problems. Please check with your instructor whether you need to know how to use it to find a limit. That said, we will need it later to prove an important theorem, so keep it in the back of your mind!

- The Squeeze, or Sandwich, Theorem says that if a function
*f*lies between two other functions near the point of interest, and those functions have the same limit*L*at that point, then*f*has that limit there as well.

On the next screen, we’ll look at a few limits involving trig functions that, early in your Calculus studies, you simply have to memorize.

Although this has been a brief discussion, if you have questions or comments about the Squeeze Theorem we’d love to hear them over on the Forum!