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Proof of the Squeeze Theorem

See how to turn eventual inequalities into an epsilon–N proof, and how the same method works when the interval between the bounds shrinks to zero.

Intermediate 9 min read

What You'll Learn

  • Translate a desired limit into two inequalities within an epsilon-neighborhood
  • Choose thresholds that make both bounding sequences close to the same limit
  • Combine eventual inequalities with convergence thresholds in a complete proof
  • Apply the proof method to rational and oscillating sequences
  • Prove a vanishing-width extension when only one bound’s limit is known
  • Check why eventual bounds and a shared limit are essential

Turning Bounds into an Epsilon–N Proof

The previous tutorial stated the Squeeze Theorem and explained how to use it. Here we prove it directly from the epsilon–N definition of convergence. The central task is to show that once both bounds lie close to their shared limit, every value between them lies close to that limit as well.

Suppose the proposed limit is \(L\), and let \(\varepsilon>0\). The epsilon–N definition asks us to find an index after which \(|x_n-L|<\varepsilon\). A useful way to establish this absolute-value inequality is to place \(x_n\) strictly between \(L-\varepsilon\) and \(L+\varepsilon\). Convergence of the lower and upper bounds supplies the two needed comparisons; eventual ordering places \(x_n\) between them.

Theorem (Squeeze Theorem): Suppose \((a_n)\), \((x_n)\), and \((b_n)\) are real sequences. Suppose there is an \(N_0\in\mathbb{N}_0\) such that
$$ a_n\leq x_n\leq b_n $$

for every \(n\geq N_0\). If \(a_n\to L\) and \(b_n\to L\), then \(x_n\to L\).

Proof. Let \(\varepsilon>0\) be arbitrary. Since \(a_n\to L\), the epsilon–N definition gives an index \(N_1\in\mathbb{N}_0\) such that

$$ |a_n-L|<\varepsilon $$

whenever \(n\geq N_1\). In particular, \(L-\varepsilon<a_n<L+\varepsilon\) for those indices. Since \(b_n\to L\), there is an index \(N_2\in\mathbb{N}_0\) such that \(L-\varepsilon<b_n<L+\varepsilon\) whenever \(n\geq N_2\). Let

$$ N=\max\{N_0,N_1,N_2\}. $$

For every \(n\geq N\), all three required conditions hold. In particular, the eventual bounds and the bounds’ proximity to \(L\) give

$$ L-\varepsilon<a_n\leq x_n\leq b_n<L+\varepsilon. $$

Therefore \(L-\varepsilon<x_n<L+\varepsilon\), which is equivalent to \(|x_n-L|<\varepsilon\). We have found an index \(N\) for each \(\varepsilon>0\) such that this inequality holds for every \(n\geq N\). By the epsilon–N definition, \(x_n\to L\). \(\square\)

Why the Thresholds Must Be Combined

The proof has three independent requirements: the lower bound must be close to \(L\), the upper bound must be close to \(L\), and the inequalities enclosing \(x_n\) must already be valid. Their thresholds need not be equal. Taking their maximum ensures that all three conditions hold at once. This is an application of the Combining Finitely Many Thresholds Theorem established earlier in the course.

Notice also that the proof uses strict inequalities at the outer edges, even though the assumed bounds may be weak. Convergence gives \(L-\varepsilon<a_n\) and \(b_n<L+\varepsilon\), while the squeeze supplies \(a_n\leq x_n\leq b_n\). Together these yield the strict inequality needed for the definition of convergence. It is not necessary for \(x_n\) to differ from either bound.

Worked Example: A Rational Sequence Trapped Near One

For \(n\in\mathbb{N}_0\), let

$$ x_n=\frac{n^2}{n^2+n+2}. $$

The denominator is positive because \(n^2\geq0\), \(n\geq0\), and the final term is \(2\). Thus \(x_n\geq0\). To compare \(x_n\) with \(1\), calculate

$$ 1-x_n =\frac{n^2+n+2-n^2}{n^2+n+2} =\frac{n+2}{n^2+n+2}. $$

This difference is nonnegative, so \(x_n\leq1\). For \(n\geq1\), the denominator satisfies \(n^2+n+2\geq n^2\), and \(n+2\leq3n\). Consequently,

$$ 0\leq1-x_n =\frac{n+2}{n^2+n+2} \leq\frac{n+2}{n^2} \leq\frac{3}{n}. $$

Equivalently, for every \(n\geq1\),

$$ 1-\frac{3}{n}\leq x_n\leq1. $$

The lower bound tends to \(1\), since \(3/n\to0\), and the upper bound is the constant sequence with value \(1\). The Squeeze Theorem gives \(x_n\to1\). The restriction \(n\geq1\) is appropriate: the estimate uses division by \(n\), so it is not asserted at \(n=0\). The original sequence is nevertheless defined there, with \(x_0=0/2=0\); that single initial term does not affect the eventual inequalities or the limit.

A Direct Error Estimate

The proof also suggests a useful way to organize estimates: instead of bounding \(x_n\) itself, one can bound how far it is from the proposed limit. The following example uses the theorem’s two-sided enclosure explicitly.

Worked Example: An Oscillating Error Around Three

Define

$$ x_n=3+\frac{(-1)^{n(n+1)/2}}{(n+2)^2}, \qquad n\in\mathbb{N}_0. $$

The integer \(n(n+1)/2\) is an exponent, so \((-1)^{n(n+1)/2}\) equals either \(-1\) or \(1\). Also \((n+2)^2>0\). Hence the error from \(3\) satisfies

$$ -\frac{1}{(n+2)^2} \leq x_n-3 \leq\frac{1}{(n+2)^2}. $$

Adding \(3\) throughout gives bounds for \(x_n\):

$$ 3-\frac{1}{(n+2)^2} \leq x_n \leq3+\frac{1}{(n+2)^2}. $$

The reciprocal-square terms tend to zero; for example, \(0\leq1/(n+2)^2\leq1/(n+1)\), and \(1/(n+1)\to0\). Thus both bounding sequences tend to \(3\). The Squeeze Theorem gives \(x_n\to3\), even though the sign of the error can change. At \(n=0\), the exponent is \(0\), so \(x_0=3+1/4\), which lies between \(3-1/4\) and \(3+1/4\). At \(n=1\), the exponent is \(1\), so \(x_1=3-1/9\), which lies between \(3-1/9\) and \(3+1/9\). These checks agree with the bounds for both signs.

When the Bounds Close Without a Shared Known Limit

The Squeeze Theorem is often applied when both bounding sequences are already known to converge to the same number. There is another useful situation: one endpoint has a known limit, and the width of the interval between the bounds tends to zero. Then the other endpoint must approach the same number, and the enclosed sequence does too. The next result proves this directly, without assuming in advance that the upper bound converges.

Theorem (Squeeze with Vanishing Width): Suppose there is an \(N_0\in\mathbb{N}_0\) such that \(a_n\leq x_n\leq b_n\) for every \(n\geq N_0\). If \(a_n\to L\) and \(b_n-a_n\to0\), then \(x_n\to L\).

Proof. Let \(\varepsilon>0\). Since \(a_n\to L\), there is an \(N_1\) such that \(|a_n-L|<\varepsilon/2\) whenever \(n\geq N_1\). Since \(b_n-a_n\to0\), there is an \(N_2\) such that \(|b_n-a_n|<\varepsilon/2\) whenever \(n\geq N_2\). For \(n\geq N_0\), the assumed ordering gives \(0\leq x_n-a_n\leq b_n-a_n\). Thus, for \(n\geq\max\{N_0,N_1,N_2\}\), the triangle inequality gives

$$ |x_n-L| =|(x_n-a_n)+(a_n-L)| \leq|x_n-a_n|+|a_n-L| \leq b_n-a_n+|a_n-L| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. $$

This is the epsilon–N condition for \(x_n\to L\), proving the result. \(\square\)

Worked Example: A Shrinking Interval Around a Limit

For \(n\in\mathbb{N}_0\), set

$$ a_n=2-\frac{2}{n+1}, \qquad b_n=2+\frac{1}{n+1}, \qquad x_n=2+\frac{(-1)^n}{n+2}. $$

The lower sequence tends to \(2\), since \(2/(n+1)\to0\). The width of the interval is

$$ b_n-a_n =\left(2+\frac{1}{n+1}\right)-\left(2-\frac{2}{n+1}\right) =\frac{3}{n+1}, $$

which tends to zero. We check the enclosure. Because \(1/(n+2)\leq2/(n+1)\) for \(n\geq0\), as follows by multiplying by the positive quantity \((n+1)(n+2)\) to get \(n+1\leq2n+4\), we have

$$ x_n\geq2-\frac{1}{n+2}\geq2-\frac{2}{n+1}=a_n. $$

Also \(1/(n+2)\leq1/(n+1)\), so

$$ x_n\leq2+\frac{1}{n+2}\leq2+\frac{1}{n+1}=b_n. $$

Both comparisons hold for every \(n\geq0\), regardless of the sign of \((-1)^n\). The Squeeze with Vanishing Width Theorem therefore gives \(x_n\to2\). For instance, at \(n=0\), the values are \(a_0=0\), \(x_0=5/2\), and \(b_0=3\), so \(a_0\leq x_0\leq b_0\). At \(n=1\), they are \(a_1=1\), \(x_1=5/3\), and \(b_1=5/2\), again in the required order.

What Can Go Wrong

The shared limit in the Squeeze Theorem cannot be omitted. The bounds \(0\leq x_n\leq1\) do not force \(x_n\) to converge: choosing \(x_n=0\) gives a sequence converging to \(0\), while choosing \(x_n=1\) gives one converging to \(1\). The interval between the bounds does not shrink around a single value.

Nor is one-sided control enough. The inequality \(x_n\leq0\) holds for the constant sequence \(x_n=-1\), which does not converge to zero. For an application, verify the direction of both inequalities, identify an index after which they hold, and confirm that the bounds approach the same value—or, for the vanishing-width result, that one bound converges and the width tends to zero. Dividing or multiplying an inequality by an expression of unknown sign can reverse its direction, so check the sign before using such a step.

1
Fix the proposed limit and tolerance.
Begin with an arbitrary \(\varepsilon>0\) and identify the neighborhood from \(L-\varepsilon\) to \(L+\varepsilon\).
2
Obtain the bounds’ thresholds.
Use convergence of the bounding sequences to place them inside that neighborhood.
3
Combine all eventual conditions.
Take the maximum of the bounds’ thresholds and the index from which the enclosure holds.
4
Conclude the epsilon inequality.
Use the enclosure to show \(L-\varepsilon<x_n<L+\varepsilon\) for every index beyond the combined threshold.

Check Your Understanding

Use the proof structure and the examples to answer these questions.

  1. In the proof of the Squeeze Theorem, why must the final threshold account for the index from which \(a_n\leq x_n\leq b_n\) holds?
  2. Why do convergence of both bounds to \(L\) and eventual enclosure imply \(|x_n-L|<\varepsilon\) eventually?
  3. For the sequence \(x_n=n^2/(n^2+n+2)\), why is it valid to use the estimate involving \(3/n\) only for \(n\geq1\)?
  4. State the additional hypotheses in the Squeeze with Vanishing Width Theorem that allow one endpoint’s limit to determine the limit of \(x_n\).
  5. Do the bounds \(0\leq x_n\leq1\) alone imply that \(x_n\) converges? Explain briefly.

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