# 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 \(\lim_{x\rightarrow 0} x^2 \sin\frac{1}{x} = 0.\)

To the right, the graph of the function under study is in red.

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## 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 and Solving Limits that Require Squeeze Theorem

**1. Upper and lower bounds explicitly given**

When the upper bound and the lower bound are explicitly given in the problem, you just need to transform the *squeezed* one into the one you're trying to find the value of.

Given an infinite sequence \(\{a_n\}\) that satisfies \(\frac{2n^2-7}{4n+5}<a_n<\frac{3n^2+8}{6n-1}\) for all positive integers \(n,\) evaluate

\[ \lim_{n\to\infty} \frac{3na_n}{(n+1)^2}.\]

Note that sending \(n\) to infinity without transforming the inequality would result in a catastrophe.We will transform the middle term of the inequality into the desired expression:

\[\frac{3n}{(n+1)^2}\cdot \frac{2n^2-7}{4n+5}<\frac{3na_n}{(n+1)^2}<\frac{3n}{(n+1)^2}\cdot\dfrac{3n^2+8}{6n-1}.\]

Now we can send \(n\) to infinity to obtain

\[\begin{align} \lim_{n\to\infty} \frac{3n(2n^2-7)}{(4n+5)(n+1)^2}\le \lim_{n\to\infty} \frac{3na_n}{(n+1)^2} &\le \lim_{n\to\infty} \frac{3n(3n^2+8)}{(6n-1)(n+1)^2} \\ \\ \implies \frac{3}{2}\le \lim_{n\to\infty} \frac{3na_n}{(n+1)^2} &\le \frac{3}{2}. \end{align}\]

Hence we conclude that

\[\lim_{n\to\infty} \frac{3na_n}{(n+1)^2} = \frac{3}{2}.\ _\square\]

This method is often used in sequences, and this type of problems is well-known in high school.

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**2. Absolute value** *(usually the function value heads to 0)*

There are many things that you could try when you see an absolute value in a problem, and the squeeze theorem is one of them.

Usually, the start is surrounding the desired expression by utilizing the properties of an absolute value.

There is a function \(f:\mathbb{R}\to\mathbb{R}.\) Prove that if \(\displaystyle \lim_{x\to a} |f(x)|=0,\) then

\[\lim_{x\to a} f(x)=0.\]

When dealing with this problem, you need to understand the following property of an absolute value:\[\text{For reals } x,~\text{it is always true that }-|x|\le x \le |x|.\]

This can be easily checked via letting \(x>0\) and \(x<0\) for respective situations.

Then, we utilize this property to surround the desired expression:

\[-|f(x)|\le f(x)\le |f(x)|.\]

Note that

\[\lim_{x\to a}\big(-|f(x)|\big)=\lim_{x\to a} |f(x)|=0.\]

This is enough to conclude that

\[\lim_{x\to a} f(x)=0.\ _\square\]

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**3. Trigonometric function (I)**

Sometimes it dazzles the student when a weird problem with trigonometric functions in limits or sums arises like the following:

Evaluate the limit

\[\lim_{n\to \infty} \sqrt[n]{3^n+\big\{2|\sin(n^n)|\big\}^n}.\]

Note that\[0\le |\sin(n^n)|\le 1.\]

So, we can write

\[\sqrt[n]{3^n} \le\sqrt[n]{3^n+\big\{2|\sin(n^n)|\big\}^n}\le \sqrt[n]{3^n+2^n}\le \sqrt[n]{2\cdot 3^n}.\]

The leftmost side is automatically \(3,\) while

\[\lim_{n\to\infty} \sqrt[n]{2\cdot 3^n}= 3 \lim_{n\to\infty} \sqrt[n]{2} = 3.\]

This is sufficient to conclude that

\[\lim_{n\to\infty} \sqrt[n]{3^n+\big\{2|\sin(n^n)|\big\}^n} = 3.\ _\square\]

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**4. Sum with a limit that tends to infinity**

When you see a sum within a limit that tends to infinity, sometimes it is wise to use the squeeze theorem.

Evaluate the limit

\[\lim_{n\to\infty} \frac{1}{n^2}\sum_{k=1}^{n} \sqrt{k^2+n+k}.\]

We observe that \(k\le n,\) so that \(\sqrt{k^2+n+k}\le \sqrt{k^2+2n}.\)We also note that for \(a>0,~b>0\) it is always true that \(\sqrt{a+b}\le \sqrt{a}+\sqrt{b}.\)

Then, we proceed to find the upper bound and the lower bound:

\[\sqrt{k^2}<\sqrt{k^2+n+k}\le \sqrt{k^2+2n} \le \sqrt{k^2}+\sqrt{2n}.\]

Now we can have

\[\frac{1}{n^2}\sum_{k=1}^{n} k<\frac{1}{n^2}\sum_{k=1}^{n} \sqrt{k^2+n+k}\le \frac{1}{n^2}\sum_{k=1}^{n} \big(k+\sqrt{2n}\big).\]

The leftmost expression converges to \(\frac{1}{2}:\)

\[\lim_{n\to\infty} \frac{1}{n^2}\sum_{k=1}^{n} k = \lim_{n\to\infty} \frac{n(n+1)}{2n^2} =\frac{1}{2}.\]

The rightmost expression also converges to \(\frac{1}{2}:\)

\[\begin{align} \lim_{n\to\infty} \frac{1}{n^2}\sum_{k=1}^{n} \big(k+\sqrt{2n}\big) &= \lim_{n\to\infty} \frac{1}{n^2}\sum_{k=1}^{n} k + \lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^{n} \sqrt{2n} \\ &= \frac{1}{2}+\lim_{n\to\infty} \frac{n\sqrt{2n}}{n^2} \\ &= \frac{1}{2}. \end{align}\]

This is sufficient to conclude that

\[\lim_{n\to\infty} \frac{1}{n^2}\sum_{k=1}^{n} \sqrt{k^2+n+k} = \frac{1}{2}.\ _\square\]

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**5. Trigonometric function (II)**

It often is the case that constant bounds just aren't enough. We then would need to find ourselves some brand-new bounds using trigonometric identities and geometry.

An infinite sequence \(\{a_n\}~(n\ge 1)\) satisfies

\[\sin\frac{1}{n+1}<a_n<\sin\frac{1}{n}\]

for all positive integers \(n.\) Prove that

\[\lim_{n\to\infty} na_n = 1.\]

Think \(x\) as \(\frac{1}{n}\) here. We can have this because \(\frac{1}{n}<\frac{\pi}{2}\) for all \(n.\)

Since the area of sector \(\rm ABD\) is in between the areas of \(\triangle \rm AEB\) and \(\triangle \rm ADF,\) we can have

\[\frac{1}{2}\sin \frac{1}{n}< \frac{1}{2n}<\frac{1}{2}\tan\frac{1}{n}.\]

Manipulating this inequality yields

\[\cos\frac{1}{n}<n\sin\frac{1}{n}<1.\]

Noting that

\[\lim_{n\to\infty} \cos \frac{1}{n}=1,\]

we can say that

\[\lim_{n\to\infty}n\sin \frac{1}{n}=1.\]

Now we need to observe this:

\[\begin{align} \lim_{n\to\infty}(n+1)\sin\frac{1}{n+1} &= 1 \\ \\ \Rightarrow \lim_{n\to\infty} n\sin\frac{1}{n+1}+\lim_{n\to\infty} \sin\frac{1}{n+1}&=1 \\ \\ \Rightarrow \lim_{n\to\infty} n\sin\frac{1}{n+1} &= 1. \end{align}\]

From the original inequality,

\[n\sin\frac{1}{n+1}<na_n<n\sin\frac{1}{n},\]

where the leftmost and rightmost expressions both approach \(1\) as \(n\) tends to infinity. Therefore,

\[\lim_{n\to\infty} na_n = 1.\ _\square\]

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**6. Functional equations**

In this type of problems, you will need some clever manipulation and/or substitution to figure out what the upper bound and the lower bound are.

Given a function \(f:\mathbb{R}\to\mathbb{R}\) that satisfies

\[f(x)-f(y)\le \left(\sqrt{|x-y|}\right)^3\]

for all reals \(x\) and \(y,\) prove that \(f\) is a constant function.

Interchange \(x\) and \(y\) to obtain\[\begin{align} f(y)-f(x)\le \left(\sqrt{|y-x|}\right)^3 \\ \\ \Rightarrow -\left(\sqrt{|x-y|}\right)^3 \le f(x)-f(y), \end{align}\]

so we can see that

\[-\left(\sqrt{|x-y|}\right)^3 \le f(x)-f(y)\le \left(\sqrt{|x-y|}\right)^3.\qquad (1)\]

Set \(x>y\) and divide both sides by \(x-y\):

\[-\sqrt{|x-y|}\le \frac{f(x)-f(y)}{x-y} \le \sqrt{|x-y|}.\]

Set \(x<y\) and divide both sides of \((1)\) by \(x-y\):

\[\sqrt{|x-y|}\ge \frac{f(x)-f(y)}{x-y} \ge -\sqrt{|x-y|}.\]

The two inequalities are exactly the same, so we proceed to the next step:

\[\lim_{y\to x} \left(-\sqrt{|x-y|}\right) = 0,\quad \lim_{y\to x} \sqrt{|x-y|} = 0.\]

So,

\[\lim_{y\to x} \frac{f(x)-f(y)}{x-y} = 0.\]

Therefore, \(f'(x)=0\) for all reals \(x,\) which proves that \(f\) is a constant function. \(_\square\)

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**7. Miscellaneous**

Sometimes there just isn't any pattern.

An infinite sequence \(\{a_n\}\) with positive terms satisfies

\[a_{n+1} = 0.9 e^{-|\tan n|} a_n\]

for all positive integers \(n.\) Prove that \(a_n\) approaches \(0\) as \(n\) tends to infinity.

It is crucial to note that the value of \(e^{-|x|}\) is always within the range \((0,~1].\)Since

\[a_n = \prod_{k=1}^{n-1} 0.9 e^{-|\tan k|} a_1,\]

we can see that

\[0 < a_n \leq 0.9^{n-1}\cdot a_1,\]

and since the right side approaches \(0\) as \(n\) tends to infinity, it is obvious that

\[\lim_{n\to\infty} a_n=0.\ _\square\]

## 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

\[\lim_{x\to0}\frac{\sin x}{x}=1.\ _\square\]

Using the squeeze theorem, evaluate

\[\lim _{ x\rightarrow \infty }{ \frac { \sin { (2x) } }{ x } } \]

Note that

\[-1\le \sin { (2x) } \le 1\implies -\frac { 1 }{ x } \le \frac { \sin { (2x) } }{ x } \le \frac { 1 }{ x }. \]

As \(\frac {1}{x}\) and \(-\frac {1}{x}\) get closer and closer to \(0\) as \(x\rightarrow \infty,\)

\[\lim _{ x\rightarrow \infty }{ \left(-\frac { 1 }{ x } \right) } =\lim _{ x\rightarrow \infty }{ \frac { 1 }{ x } } =0.\]

Thus,

\[\lim _{ x\rightarrow \infty }{ \frac { \sin { (2x) } }{ x } } =0.\ _\square\]

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. \(_\square\)