# Squeeze Theorem

The **squeeze theorem** is a theorem used in calculus to evaluate a limit of a function.

The theorem is particularly useful to evaluate limits where other techniques might be unnecessarily complicated. For example, \(\lim_{x\rightarrow 0} x^2 \sin\frac{1}{x}\) is somewhat tricky as it is the product of a term that goes to 0 and one that does not converge. However, since \(-1 \le \sin\frac{1}{x} \le 1,\) \(-x^2 \le x^2 \sin\frac{1}{x} \le x^2,\) and \(-x^2\) and \(x^2\) both tend to \(0\) as \(x\rightarrow 0,\) it must be the case that \(\lim_{x\rightarrow 0} x^2 \sin\frac{1}{x} = 0.\)

In the graph of the example to the right, \(\sin\frac{1}{x}\) is in red, \(x^2\) is in green, and the function under study is in black.

#### Contents

## Statement of the Theorem and its Intuition

**Intuition**: Another name for the squeeze theorem is the **sandwich theorem**. Notice how the theorem makes a "sandwich" of function \(f\) between functions \(g\) and \(h\) on a subset of \(\mathbb{R}\). \(g\) and \(h\) have been "cleverly chosen" (if possible) to have the same limit as \(f\). Since \(g\) and \(h\) enclose \(f,\) the three functions must all have the same limit.

Assume that functions \(f,\ g,\ h\) defined in \(D \subseteq \mathbb{R}\) satisfy

\[g(x)\le f(x)\le h(x), \space \forall x \in D.\]

Then, if \(\displaystyle \lim_{x\to a} g(x) = \lim_{x \to a} h(x) = l,\) then \(\displaystyle \lim_{x \to a} f(x) = l.\)

## Proof of the Theorem

This theorem is equivalent to the next theorem: If three sequences\((a_n),(b_n),(c_n)\) fulfill \(a_n \leq b_n \leq c_n\) \(\forall n > n_{\small{0}}\) with \(n, n_{\small{0}} \in \mathbb{N}\) and \((a_n),(c_n)\) have the same limit \(l\), then \((b_n)\) has the same limit \(l\), due to \(\displaystyle \lim_{x \to a} f(x) = l\) if and only if \(\forall (x_n) \to a, \space x_n \neq a, x_n \in D\) is \(\displaystyle \lim_{n \to \infty} f(x_n) = l\).

We are going to prove the last theorem: Given an \( \epsilon > 0, \space \exists n_{\small{0}} \space / \space \forall n > n_{\small{0}}, \space n \in \mathbb{N},\space a_n, \space c_n \in (l - \epsilon, l + \epsilon) \implies b_n \in (l - \epsilon, l + \epsilon) \space \forall n > n_{\small{0}},\) which implies \(\displaystyle \lim_{n \to \infty} b_n = l.\) \(_\square\)

## Identifying Limits that require Squeeze Theorem

## Determining Appropriate Lower and Upper Bounds

## Examples

Evaluate \[\lim_{n\to\infty}{\dfrac{n^{k}\sin^{2}\left(n!\right)}{n+2}}.\]

Since \(0 \le \sin^2 x \le 1 \ \forall x \in \mathbb R,\) \[ 0 \le \dfrac {n^k\sin^2 (n!)}{n+2} \le \dfrac {n^k}{n+2}.\] Note that \(\lim_{n \to \infty} \frac {n^k}{n+2} = 0\) for \(k \in (0, 1),\) which implies \[0 \le \displaystyle \lim_{n \to \infty} \dfrac {n^k\sin^2 (n!)}{n+2} \le 0.\] By the squeeze theorem, we get \[\lim_{n \to \infty} \dfrac {n^k\sin^2 (n!)}{n+2} = 0.\ _\square\]

Evaluate \[\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+\sqrt2}+\cdots+\frac{1}{n+\sqrt n}\right).\]

Let \(S_n=\frac{1}{n+1}+\frac{1}{n+\sqrt2}+\cdots+\frac{1}{n+\sqrt n},\) then \[\frac{n}{n+\sqrt n}<S_n<\frac{n}{n+1}.\] Note that \(\lim_{n\to\infty}\frac{n}{n+\sqrt n}=\lim_{n\to\infty}\frac{n}{n+1}=1,\) which implies \[\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+\sqrt2}+\cdots+\frac{1}{n+\sqrt n}\right)=1.\ _\square\]

Evaluate \[\lim_{n\to\infty}\left(\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n}\right).\]

Let \(S_n=\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n},\) then \[\begin{align} \frac{1}{n^2+n+n}+\frac{2}{n^2+n+n}+\cdots +\frac{n}{n^2+n+n} &<\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n}\\ &<\frac{1}{n^2+n+1}+\frac{2}{n^2+n+1}+\cdots +\frac{n}{n^2+n+1}\\ \Rightarrow \frac{\frac{1}{2}n(n+1)}{n^2+n+n}&<S_n<\frac{\frac{1}{2}n(n+1)}{n^2+n+1}. \end{align}\] Note that \[\lim_{n\to\infty}\frac{\frac{1}{2}n(n+1)}{n^2+n+n}=\lim_{n\to\infty}\frac{\frac{1}{2}n(n+1)}{n^2+n+1}=\frac{1}{2},\] which implies \[\lim_{n\to\infty}\left(\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n}\right)=\frac{1}{2}.\ _\square\]

Use the squeeze theorem to prove that \( \displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

When \(x\) is close to 0, we have \[\cos x<\frac{\sin x}{x}<1.\] Note that \(\lim_{x\to0}\cos x=\lim_{x\to0}1=1,\) which implies \[\lim_{x\to0}\frac{\sin x}{x}=1.\ _\square\]

Evaluate \[ \lim_{n\to\infty} \sum_{k=1}^n \dfrac{1}{\sqrt{kn}}.\]

We know that \(\sqrt{k-1}+\sqrt{k}<2\sqrt{k}<\sqrt{k}+\sqrt{k+1},\) which implies \[\begin{align} 2\left(\sqrt{k+1}-\sqrt{k}\right)<\dfrac{1}{\sqrt{k}}&<2\left(\sqrt{k}-\sqrt{k-1}\right) \\ 2\sum_{k=1}^n \left(\sqrt{k+1}-\sqrt{k}\right)<\sum_{k=1}^n \dfrac{1}{\sqrt{k}}&<2\sum_{k=1}^n \left(\sqrt{k}-\sqrt{k-1}\right) \\ 2\sqrt{n+1}-2<\sum_{k=1}^n \dfrac{1}{\sqrt{k}}&<2\sqrt{n} \\ \dfrac{2\sqrt{n+1}-2}{\sqrt{n}}<\sum_{k=1}^n \dfrac{1}{\sqrt{kn}}&<\dfrac{2\sqrt{n}}{\sqrt{n}}=2. \end{align} \] Note that \(\lim_{n\to\infty}\frac{2\sqrt{n+1}-2}{\sqrt{n}}=2,\) so \[\lim_{n\to\infty} \sum_{k=1}^n \dfrac{1}{\sqrt{kn}}=2.\ _\square\]

Find \(\displaystyle \lim_{t \to 0} \frac{\sin t}t. \)

We start with the inequality \(\sin t ≤t≤ \tan t\) that is pretty much the definition of the radian measure of the angles. Dividing by \(t\) yields \(\cos t ≤\frac{ \sin t}t ≤ 1.\) But \(\cos t = 1-2\sin^2 \frac t2 ≥ 1-\frac{t^2}2\) and we can apply the squeeze theorem. We assumed \(t\) to be positive, but \( \frac{\sin t}t \) is even, so we are done. Actually, we could simply take \(\delta = \sqrt{2 \epsilon} \) instead of using the squeeze theorem to finish the proof. \(_\square\)