When Direct Evaluation Is Difficult
A sequence can have terms that oscillate or combine several effects, making its limit hard to determine by evaluating the formula directly. Sometimes the useful information is not an exact expression for each term, but a pair of inequalities that trap every term between two simpler sequences.
If the lower and upper sequences approach the same number, the trapped sequence has no room to approach anything else. This idea is the Squeeze Theorem for sequences. It applies even when the sequence in the middle is not monotonic and its terms fluctuate from one index to the next.
The Theorem and Its Conditions
For a sequence \((x_n)\), the statement \(x_n\to L\) means that for every \(\varepsilon>0\), there is a positive integer \(N\) such that \(n\ge N\) implies \(|x_n-L|<\varepsilon\). In words, after some index, every term lies within any chosen positive distance of \(L\).
The inequalities need only hold eventually: a finite number of early terms do not affect a sequence's limit. The lower and upper sequences must converge to the same number. The theorem does not require the middle sequence to be monotonic or to converge in advance; convergence is the conclusion.
In practice, the work is usually to find useful bounds, prove that they really enclose the sequence, and show that both bounds approach one common limit. Limit laws for sequences can then be used to evaluate the limits of the simpler bounding sequences.
Why the Theorem Works
The proof follows directly from the definition of a sequence limit. Take any desired error tolerance \(\varepsilon>0\). Since \(a_n\to L\), eventually \(a_n>L-\varepsilon\). Since \(c_n\to L\), eventually \(c_n<L+\varepsilon\). From some index onward, the middle term is therefore also between \(L-\varepsilon\) and \(L+\varepsilon\).
Let \(\varepsilon>0\) be arbitrary. The aim is to establish \(|x_n-L|<\varepsilon\) for all sufficiently large \(n\).
Because \(a_n\to L\), choose \(N_1\) so \(n\ge N_1\) gives \(|a_n-L|<\varepsilon\), and hence \(a_n>L-\varepsilon\). Because \(c_n\to L\), choose \(N_2\) so \(n\ge N_2\) gives \(|c_n-L|<\varepsilon\), and hence \(c_n<L+\varepsilon\).
For \(n\ge \max(N_0,N_1,N_2)\), the hypotheses give \(L-\varepsilon<a_n\le x_n\le c_n<L+\varepsilon\).
The preceding chain implies \(-\varepsilon<x_n-L<\varepsilon\), so \(|x_n-L|<\varepsilon\). Since this holds for every \(\varepsilon>0\), \(x_n\to L\).
This proof explains why the two limits must agree. If the bounds approach different values, they may leave room for the middle sequence to keep varying. Matching limits squeeze that room down to zero.
Worked Example: An Alternating Sequence Approaches Zero
Worked Example: An Alternating Sequence Approaches Zero
Find the limit of \(x_n=\frac{(-1)^n}{n+1}\), where \(n\in\mathbb{N}\).
Since \((-1)^n\) is either \(-1\) or \(1\), it always satisfies \(-1\le (-1)^n\le 1\). As \(n+1>0\), dividing all parts by \(n+1\) preserves the inequality:
The lower and upper sequences have limits
because \(1/(n+1)\to 0\). The Squeeze Theorem therefore gives \(\lim_{n\to\infty}x_n=0\). Check: the inequalities trap every term between quantities whose distance from zero is \(1/(n+1)\), and that distance tends to zero, as required.
The signs of the terms keep changing, but the magnitude of each term is at most \(1/(n+1)\). The theorem uses that shrinking bound; it does not require the alternating terms themselves to move consistently in one direction.
Worked Example: A Bounded Oscillation Divided by \(n\)
Worked Example: A Bounded Oscillation Divided by \(n\)
Find the limit of \(x_n=\frac{\sin(n^2)}{n}\), where the sine input is measured in radians.
For every real number \(t\), \(-1\le\sin(t)\le 1\). In particular, this holds for \(t=n^2\). Since \(n>0\), division by \(n\) gives
The two bounds have the same limit:
It follows from the Squeeze Theorem that \(\lim_{n\to\infty}\frac{\sin(n^2)}{n}=0\). Check: the estimate \(\left|\frac{\sin(n^2)}{n}\right|\le\frac{1}{n}\) follows from \(|\sin(n^2)|\le1\), and its right-hand side tends to zero. The sine values may oscillate, but their contribution is divided by an increasingly large \(n\).
Notice what the argument does not do: it does not try to calculate or predict the individual values \(\sin(n^2)\). The uniform bound \(-1\le\sin(n^2)\le1\) is enough.
Worked Example: An Oscillating Perturbation Around a Nonzero Limit
Worked Example: An Oscillating Perturbation Around a Nonzero Limit
Find the limit of \(x_n=\frac{3n+(-1)^n}{5n+2}\).
Set \(e_n=(-1)^n\), so \(-1\le e_n\le1\). First compare \(x_n\) with \(3/5\). Since \(5n+2>0\),
The bound \(-1\le e_n\le1\) gives \(-11\le5e_n-6\le-1\), and therefore \(|5e_n-6|\le11\). Because \(25n+10>0\), it follows that
Adding \(3/5\) to all three parts gives the required bounds on \(x_n\):
Both bounding sequences approach \(3/5\), since \(11/(25n+10)\to0\). Thus the Squeeze Theorem gives \(\lim_{n\to\infty}x_n=3/5\). Check: subtracting \(3/5\) from the original expression produces exactly \((5e_n-6)/(25n+10)\), whose absolute value is at most \(11/(25n+10)\to0\). Hence the difference from \(3/5\) tends to zero.
This example illustrates a useful strategy: when a sequence consists of a main part plus an oscillating part, estimate the size of the oscillation and show that its effect becomes arbitrarily small. The limit need not be zero; the two bounds can instead converge to any common value.
A Common Mistake: Using a Bound That Is Not True
A squeeze argument is only valid if the inequalities hold for every sufficiently large index. For example, it is incorrect to claim
The middle term is negative whenever \(n\) is odd. At \(n=1\), it equals \(-1\), so the claimed lower bound \(0\le-1\) is false. Because the inequality fails for infinitely many indices, it cannot be repaired by ignoring a finite number of early terms.
The valid estimate is
Both outer sequences tend to zero, so this corrected squeeze proves that \(\frac{(-1)^n}{n}\to0\). Before applying the theorem, check the signs of denominators when dividing inequalities and verify that the proposed bounds really contain every relevant term.
Using the Squeeze Theorem Reliably
When a sequence is difficult to evaluate directly, organize the argument around three checks:
- Enclosure: establish \(a_n\le x_n\le c_n\) for all sufficiently large \(n\).
- Matching limits: verify that both bounds approach the same number \(L\).
- Conclusion: state that the Squeeze Theorem gives \(x_n\to L\).
Typical useful bounds come from a known range, such as \(-1\le\sin(t)\le1\), or from an estimate showing that an oscillating term has small magnitude. Bounds that are easy to write but do not have a common limit are not enough. Nor is it enough to observe that several terms appear close to a number: the inequalities and the limits of both bounding sequences must justify the conclusion.
Check Your Understanding
For each question, identify the bounds or condition that makes the conclusion valid.
- If \(-1\le\cos(n)\le1\), what limit follows for \(\frac{\cos(n)}{n}\), and what two sequences provide the squeeze?
- Why is it enough for \(a_n\le x_n\le c_n\) to hold for all \(n\ge N_0\), rather than for every positive integer \(n\)?
- Suppose \(a_n\le x_n\le c_n\), \(a_n\to2\), and \(c_n\to2\). What does the Squeeze Theorem conclude?
- Why do the bounds \(-1\le(-1)^n\le1\) alone not prove that \((-1)^n\) has a limit?
- For \(x_n=\frac{4n+(-1)^n}{7n+1}\), what common value should bounds around \(x_n\) approach?
