When Bounds Determine a Limit
Limit laws let us find the limit of a sequence by combining limits already known. Sometimes, however, a sequence is difficult to analyze directly even though it is easy to place between two simpler sequences. If the simpler sequences approach the same number, their bounds leave less and less room for the sequence between them. The Squeeze Theorem formalizes this idea.
The bounds need not hold at every index. Convergence is determined by the terms from some index onward, so it is enough for the inequalities to hold eventually: there must be an index \(N_0\in\mathbb{N}_0\) such that they hold for every \(n\geq N_0\). We will also establish a useful fact about eventual inequalities between convergent sequences, then use the Squeeze Theorem in examples.
Eventual Inequalities and Limits
Proof. Suppose, for contradiction, that \(L>M\). Then \(M<L\). Apply the Eventual Ordering from Distinct Limits Theorem, established earlier in the course, to the convergent sequences \((b_n)\) and \((a_n)\). Since their limits satisfy \(M<L\), that theorem gives an \(N_1\) such that
for every \(n\geq N_1\). For every \(n\geq\max\{N_0,N_1\}\), we would therefore have both \(a_n\leq b_n\) and \(b_n<a_n\), which is impossible. Thus the supposition \(L>M\) is false, and \(L\leq M\). \(\square\)
The conclusion is not necessarily strict, even when the terms are strictly ordered. For example, \(a_n=1-1/(n+1)\) and \(b_n=1\) satisfy \(a_n<b_n\) for every \(n\), but both sequences converge to \(1\). Eventual order is preserved as a weak inequality between the limits.
for every \(n\geq N_0\). Then \(x_n\to L\).
Proof. Let \(\varepsilon>0\). Since \(u_n\to0\), there is an \(N_1\) such that \(u_n<\varepsilon\) whenever \(n\geq N_1\). For every \(n\geq\max\{N_0,N_1\}\), the assumed bound gives
This is exactly the epsilon–N condition for \(x_n\to L\). \(\square\)
This corollary is often a convenient way to use the squeeze idea when the quantity of interest is an error. It is enough to find a nonnegative upper bound for the absolute error that tends to zero; one does not need to calculate the sequence’s limit by manipulating its formula.
The Squeeze Theorem
for every \(n\geq N_0\). If \(a_n\to L\) and \(b_n\to L\), then \(x_n\to L\).
The lower and upper sequences can be thought of as walls enclosing \(x_n\). Once both walls are close to \(L\), every value between them must also be close to \(L\). The requirement that both limits equal \(L\) is essential to this conclusion: the two walls must close in around the same number. The proof will be developed in “Proof of the Squeeze Theorem.”
A useful special case has \(L=0\). If \(a_n\leq x_n\leq b_n\) eventually and both bounds tend to zero, then \(x_n\to0\). Another common form uses absolute values: if \(|x_n|\leq u_n\) eventually, with \(u_n\geq0\) and \(u_n\to0\), the Vanishing Error Bound Corollary gives \(x_n\to0\).
Worked Examples
Worked Example: An Alternating Sequence
Consider \(x_n=(-1)^n/(n+1)\). Since \((-1)^n\) is either \(-1\) or \(1\), for every \(n\in\mathbb{N}_0\),
The lower and upper sequences both tend to zero. For instance, the reciprocal bound \(1/(n+1)\to0\) follows from the reciprocal-bound criterion for convergence established earlier in the course. The Squeeze Theorem therefore gives
The terms alternate in sign, so the sequence is not monotone. That does not prevent convergence: the size of each term is exactly \(1/(n+1)\), which becomes arbitrarily small. At \(n=0\), the term is \(1\), and at \(n=1\), it is \(-1/2\); both values satisfy the displayed bounds. The bounds, rather than monotonicity, determine the limit.
Worked Example: A Rational Sequence Trapped Above Zero
For \(n\in\mathbb{N}_0\), define
The denominator \(n^2+n+1\) is positive, since \(n\geq0\) makes each of its three terms nonnegative and the final term is \(1\). Thus \(x_n\geq0\). To obtain an upper bound, compare \(x_n\) with \(1/(n+1)\). Both denominators are positive, so the desired inequality is equivalent to
The left side is \(n^2+n\), so this inequality holds because \(n^2+n\leq n^2+n+1\). Consequently,
Both bounds tend to zero, and hence the Squeeze Theorem gives \(x_n\to0\). The estimate also works at \(n=0\): the middle term is \(0/1=0\), which lies between \(0\) and \(1\).
Worked Example: A Sequence Converging to One
Let
To study convergence to \(1\), subtract \(1\) and combine the fractions. Since \(n+2>0\) for every \(n\geq0\),
Because \((-1)^n\) is either \(-1\) or \(1\), the numerator \((-1)^n-2\) is either \(-3\) or \(-1\). In particular, its absolute value is at most \(3\), and therefore
Both bounding sequences tend to zero. The Squeeze Theorem gives \(x_n-1\to0\), which means \(x_n\to1\). The algebra can be checked at the first two indices: \(x_0=(0+1)/2=1/2\), so \(x_0-1=-1/2\), and the bound is \(-3/2\leq-1/2\leq3/2\). Also \(x_1=(1-1)/3=0\), so \(x_1-1=-1\), and \(-1\leq-1\leq1\). In each case the error lies between the claimed bounds.
Worked Example: Using a Bound on the Absolute Error
Suppose a sequence is given by
Rather than trap \(y_n\) directly, estimate its distance from \(4\):
The equality uses \(|(-1)^n|=1\), and the inequality holds because \(n+1\geq1\), so \((n+1)^2\geq n+1\). Since \(1/(n+1)\to0\), the Vanishing Error Bound Corollary shows that \(y_n\to4\). For example, \(y_0=5\) and \(|y_0-4|=1=1/(0+1)\); at \(n=1\), \(y_1=4-1/4=15/4\) and \(|y_1-4|=1/4\leq1/2\). The estimate applies at these indices as well as throughout the sequence.
What the Bounds Must—and Need Not—Do
The inequalities in the Squeeze Theorem need only hold eventually, not at every index. A finite number of exceptional terms cannot affect convergence, as established by the Finite Changes Preserve Convergence Theorem. In practice, this means that a denominator may require a separate check at early indices, or an estimate may become valid only after some threshold; either is acceptable if the sequence is properly defined and the inequalities hold from then on.
The shared limit is the crucial condition. For example, \(0\leq x_n\leq1\) does not determine a limit: the sequence \(x_n=0\) satisfies these bounds and converges to \(0\), while \(x_n=1\) also satisfies them and converges to \(1\). The bounds do not approach one another around a single value. Likewise, knowing only \(x_n\leq b_n\) with \(b_n\to0\) is insufficient: the constant sequence \(x_n=-1\) satisfies \(x_n\leq b_n=0\), but does not converge to zero. Both a lower and an upper bound approaching the same value are what make the squeeze effective.
When applying the theorem, check that the inequalities point in the right direction, that they hold for every sufficiently large index, and that both bounding sequences have the same limit. If the goal is to prove convergence to \(L\), it is often helpful to subtract \(L\) and bound the absolute error, as in the final worked example.
Identify the number that both bounding sequences should approach.
Verify both inequalities and check any conditions, such as positivity of denominators, needed to preserve their direction.
Confirm that both bounds tend to the same value; then invoke the Squeeze Theorem.
Check Your Understanding
Use eventual order, the Squeeze Theorem, and the worked estimates to answer these questions.
- If \(a_n\leq b_n\) for every sufficiently large \(n\), \(a_n\to L\), and \(b_n\to M\), what inequality must hold between \(L\) and \(M\)?
- Suppose \(-1/(n+1)\leq x_n\leq1/(n+1)\). What limit follows for \((x_n)\), and why?
- Why do the bounds \(0\leq x_n\leq1\) alone not determine the limit of \((x_n)\)?
- If \(|x_n-L|\leq u_n\) eventually, \(u_n\geq0\), and \(u_n\to0\), what does the Vanishing Error Bound Corollary imply?
- Does an eventual inequality \(x_n\leq b_n\), with \(b_n\to0\), by itself imply \(x_n\to0\)? Give a reason for your answer.