Prove the Squeeze Theorem

From a Unique Limit to a Trapped Limit

In “Prove Uniqueness of a Sequence Limit,” the proof compared two candidate limits by controlling their distances from the same sequence term. The Squeeze Theorem uses a related idea: instead of comparing two proposed limits, compare a sequence to two other sequences that bound it. If both bounds approach the same real number, the middle sequence has no room to approach anything else.

The essential work is keeping the inequalities valid on one common tail. The bounds need not hold for every term; finitely many exceptions do not obstruct the argument. This is why the theorem is naturally stated using eventual inequalities.

Definition: A property of the positive integers holds eventually if there is a positive integer \(N_0\) such that the property holds for every integer \(n\geq N_0\).

For example, \(a_n\leq b_n\) eventually means that there is one threshold \(N_0\) after which \(a_n\leq b_n\) for every term. The threshold is independent of \(n\) once the tail begins. Two eventual inequalities may initially have different thresholds; taking their maximum gives a tail on which both hold.

The Squeeze Theorem

Theorem (Squeeze Theorem): Let \((a_n)\), \((b_n)\), and \((c_n)\) be real sequences. Suppose \(a_n\leq b_n\leq c_n\) eventually, and suppose \(a_n\to L\) and \(c_n\to L\) for some \(L\in\mathbb R\). Then \(b_n\to L\).

Proof. Let \(\varepsilon>0\) be arbitrary. Since \(a_n\to L\), there is a positive integer \(N_a\) such that for every \(n\geq N_a\),

L-\varepsilon<a_n<L+\varepsilon.

Since \(c_n\to L\), there is a positive integer \(N_c\) such that for every \(n\geq N_c\),

L-\varepsilon<c_n<L+\varepsilon.

By the eventual-inequality hypothesis, there is a positive integer \(N_0\) such that \(a_n\leq b_n\leq c_n\) for every \(n\geq N_0\). Set \(N=\max\{N_a,N_c,N_0\}\). For every \(n\geq N\), all four bounds apply. In particular,

L-\varepsilon<a_n\leq b_n\leq c_n<L+\varepsilon.

Therefore \(L-\varepsilon<b_n<L+\varepsilon\), which is equivalent to \(|b_n-L|<\varepsilon\). Since this holds for every \(n\geq N\), and \(\varepsilon>0\) was arbitrary, the definition of convergence gives \(b_n\to L\). \(\square\)

The proof uses strict inequalities at the outer edges of the epsilon interval and allows non-strict inequalities between the three sequences. There is no need for \(a_n<b_n<c_n\): equality with either bound is harmless. What matters is that both bounds eventually fit inside the same interval around \(L\).

Identify the common target.
Check that the lower and upper sequences both converge to the same real number \(L\).

Fix an arbitrary tolerance.
Use convergence of each bounding sequence to place it between \(L-\varepsilon\) and \(L+\varepsilon\).

Choose one tail.
Take the maximum of the two convergence thresholds and the threshold for the eventual inequalities.

Trap the middle term.
Combine the inequalities to obtain \(|b_n-L|<\varepsilon\) on that tail.

Worked Applications of the Theorem

Worked Example: A Positive Rational Sequence

Let \(b_n=n/(n^2+4)\) for positive integers \(n\). We will show that \(b_n\to0\) by bounding it on both sides. Since \(n>0\) and \(n^2+4>0\), we have \(b_n>0\). Also, \(n^2+4\geq n^2\), so division by the positive quantities gives

0\leq \frac{n}{n^2+4}\leq\frac{n}{n^2}=\frac1n.

The lower sequence is constantly zero and therefore converges to zero. For the upper sequence, given \(\varepsilon>0\), choose a positive integer \(N>1/\varepsilon\). For \(n\geq N\), \(1/n\leq1/N<\varepsilon\), so \(1/n\to0\). Thus both bounds converge to zero, and the Squeeze Theorem gives \(n/(n^2+4)\to0\).

Worked Example: An Oscillating Sequence

Consider \(b_n=(-1)^n/(n+1)\). Its sign alternates, so a one-sided estimate from zero would not be sufficient. Since \(|(-1)^n|=1\) and \(n+1>0\), its absolute value is \(1/(n+1)\), and hence

-\frac1{n+1}\leq\frac{(-1)^n}{n+1}\leq\frac1{n+1}.

Both bounding sequences converge to zero. To verify this directly, let \(\varepsilon>0\) and choose a positive integer \(N>1/\varepsilon\). For every \(n\geq N\), \(0<1/(n+1)\leq1/n\leq1/N<\varepsilon\). The negative bound also has absolute value \(1/(n+1)<\varepsilon\), so it converges to zero as well. The Squeeze Theorem now shows that the oscillating sequence converges to zero. The signs do not prevent convergence because the magnitudes of the terms are trapped by quantities tending to zero.

Worked Example: A Difference of Square Roots

Let \(b_n=\sqrt{n^2+2n}-n\). Directly, this is a difference of two large quantities, so its limit is not immediately apparent. Rationalizing and dividing numerator and denominator by \(n>0\) gives

b_n =\frac{(\sqrt{n^2+2n}-n)(\sqrt{n^2+2n}+n)}{\sqrt{n^2+2n}+n} =\frac{2n}{\sqrt{n^2+2n}+n} =\frac{2}{\sqrt{1+2/n}+1}.

Since \(2/n>0\), we have \(\sqrt{1+2/n}>1\), so the last expression is positive and less than \(1\). We can also bound how far it lies below \(1\):

1-b_n =\frac{\sqrt{1+2/n}-1}{\sqrt{1+2/n}+1} =\frac{2/n}{(\sqrt{1+2/n}+1)^2} \leq\frac{1}{2n}.

For the equality, the numerator \(\sqrt{1+2/n}-1\) was rationalized. For the final inequality, \(\sqrt{1+2/n}+1\geq2\), so its square is at least \(4\). These calculations show

1-\frac{1}{2n}\leq b_n\leq1.

The lower bound converges to \(1\), because its distance from \(1\) is \(1/(2n)\), which tends to zero; the upper bound is constantly \(1\). Applying the Squeeze Theorem gives \(b_n\to1\). This example illustrates a useful proof pattern: first transform an expression into a form that makes suitable bounds visible, then use the theorem to finish the limit argument.

A Vanishing-Error Consequence

The Squeeze Theorem also explains why a sequence can be replaced by an approximation whose error tends to zero. The next result states this principle directly and proves it with the triangle inequality.

Theorem (Vanishing Error Preserves a Sequence Limit): Let \((x_n)\) and \((y_n)\) be real sequences, and let \(L\in\mathbb R\). Suppose \(y_n\to L\) and \(|x_n-y_n|\to0\). Then \(x_n\to L\).

Proof. Let \(\varepsilon>0\). Since \(y_n\to L\), there is a positive integer \(N_y\) such that \(|y_n-L|<\varepsilon/2\) for every \(n\geq N_y\). Since \(|x_n-y_n|\to0\), there is a positive integer \(N_e\) such that \(|x_n-y_n|<\varepsilon/2\) for every \(n\geq N_e\). Set \(N=\max\{N_y,N_e\}\). For \(n\geq N\), the triangle inequality for absolute values gives

|x_n-L| \leq |x_n-y_n|+|y_n-L| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.

This is the definition of \(x_n\to L\), so the result follows. \(\square\)

In practice, an error estimate often gives \(|x_n-y_n|\leq r_n\), where \(r_n\geq0\) and \(r_n\to0\). Then the vanishing-error theorem applies: for every \(\varepsilon>0\), eventually \(r_n<\varepsilon/2\), and consequently \(|x_n-y_n|<\varepsilon/2\). This lets a simpler sequence \(y_n\) stand in for a more complicated \(x_n\), provided the difference becomes negligible.

What the Bounds Do—and Do Not—Tell You

The two bounding sequences must have the same limit for the Squeeze Theorem to identify the middle sequence’s limit. For example, the inequalities \(0\leq b_n\leq1\) alone do not imply that \(b_n\) converges. The sequence that is \(0\) for even \(n\) and \(1\) for odd \(n\) satisfies those inequalities for every \(n\), yet its even and odd terms do not approach one common value. The bounds confine the sequence, but they do not narrow the allowed values to one target.

It is also important to distinguish an eventual inequality from a pointwise inequality. The theorem requires one threshold after which both comparisons hold. If the lower comparison begins at \(N_1\) and the upper comparison begins at \(N_2\), use \(N_0=\max\{N_1,N_2\}\). The proof must then also account for the thresholds from convergence of the bounds, which is why its final choice takes the maximum of three integers.

Earlier in this course, the Theorem (Order Is Preserved Under Limits) established that inequalities between convergent sequences pass to their limits. That result is useful when the middle sequence is already known to converge. The Squeeze Theorem addresses a different task: it establishes convergence of the middle sequence from the two bounds, without assuming in advance that the middle sequence has a limit. In a proof, this difference matters. Do not assume the very convergence you are trying to prove.

When searching for a squeeze, aim for bounds that are both easy to verify and known to approach the proposed limit. Absolute-value estimates often provide the needed pair: if \(|b_n-L|\leq r_n\) eventually and \(r_n\to0\), then \(L-r_n\leq b_n\leq L+r_n\), and the two bounds converge to \(L\). The proof still depends on verifying the inequalities and the limit of the error bound; naming the Squeeze Theorem does not replace those steps.

Check Your Understanding

Use the epsilon definition of convergence and the stated inequalities to answer each question.

Why does the proof of the Squeeze Theorem take the maximum of three thresholds?
Does the theorem require strict inequalities between the lower, middle, and upper sequences? Explain.
Why do the inequalities \(0\leq b_n\leq1\) alone fail to guarantee that \(b_n\) converges?
If \(|x_n-y_n|\to0\) and \(y_n\to L\), which triangle inequality gives the needed estimate for \(|x_n-L|\)?
What additional fact about the bounds is needed before the Squeeze Theorem can identify a limit for the middle sequence?

What You'll Learn