The Sandwich Theorem

There comes a time in everyone’s life where, no matter how hard we try, we just cannot evaluate \(\lim\limits_{x\rarr0}x^2\sin(\frac{1}{x^2})\). It is a gatekeeper to the echelons of calculus, yet, with the proper tools, the evaluation becomes quite simple.

Our biggest obstacle is the \(\sin(\frac{1}{x^2})\) factor. Normally,

\(\lim\limits_{x\rarr{a}}f(x)\cdot{g(x)} = \lim\limits_{x\rarr{a}}f(x) \cdot \lim\limits_{x\rarr{a}}g(x)\)

But as \(\lim\limits_{x\rarr{0}}\sin(\frac{1}{x^2})\) does not exist, we’ll have to try another method.

Introducing: The Squeeze Theorem

The Squeeze Theorem, also known as the sandwich theorem, states that on an interval where \(a\) is included, and \(f(x)\), \(g(x)\), and \(h(x)\) are defined on every point of the interval with sometimes the exception of \(a\), and \(g(x)\leq f(x) \leq h(x)\), if

\(\lim\limits_{x\rarr{a}}g(x) = L = \lim\limits_{x\rarr{a}}h(x)\)
then \(\lim\limits_{x\rarr{a}}f(x) = L\)

Essentially, we can “squeeze” a difficult function in between two “simpler” functions with known limits, and if the two outer functions have the same limit, than the inner function must also have that same limit. Now the question becomes, which functions do we squeeze \(x^2\sin(\frac{1}{x^2})\) in between?

We can start by just looking at the \(\sin(\frac{1}{x^2})\) factor. The \(\sin\) function has a range of \([-1, 1]\) which means that

\(-1 \leq \sin(\frac{1}{x^2}) \leq 1\)

From here, we can just multiply all sides of the inequality by \(x^{2}\), which results in

\(-x^2 \leq x^2\sin(\frac{1}{x^2}) \leq x^2\)

When we find the limits of the outer functions, we can see that

\(\lim\limits_{x\rarr{0}}-x^2 = 0 = \lim\limits_{x\rarr{0}}x^2\)

which, via the squeeze theorem, means that

\(\lim\limits_{x\rarr{0}}x^2\sin(\frac{1}{x^2}) = 0\)

A Desmos graph of the three equations is shown below. Notice how \(f(x)\) lies in between \(g(x)\) and \(h(x)\).

A Second Example

The first limit was pretty simple, so let’s take a look at a different one. Let’s consider the following limit:

\(\lim\limits_{x\rarr{\infty}}\frac{\cos^2({5x})}{3-2x}\)

\(\lim\limits_{x\rarr{\infty}}\cos^2({5x})\) does not exist, but we can use the squeeze theorem to find this limit. First, we know that

\(0 \leq \cos^{2}(5x) \leq 1 \) because the range of \(\cos^2(5x)\) is \([0, 1]\).

Now we can divide all sides of the inequality by \(3-2x\). Since \(x\rarr\infty\), \(3-2x\) will be negative, which means our inequalties will switch. Therefore,

\(\frac{0}{3-2x} \geq \cos^{2}(5x) \geq \frac{1}{3-2x} \)

\(= \frac{1}{3-2x} \leq \cos^{2}(5x) \leq 0\)

We can take a look at the limits of the outside functions:

\(\lim\limits_{x\rarr{\infty}}\frac{1}{3-2x} = 0 = \lim\limits_{x\rarr{\infty}}0\)

which, via the squeeze theorem, means that

\(\lim\limits_{x\rarr{\infty}}\frac{cos^2(5x)}{3-2x} = 0\)

Conclusion

Whether your sandwich is made of meat and bread or functions and functions, we hope that you can make at least some use of the squeeze theorem. It’s not always useful—but when it is, there’s probably a couple of other methods you could use instead.