Proofs with Maxima and Minima
A useful proof often begins by rewriting a complicated expression in a form that the available limit laws can handle. The maximum or minimum of two terms is a good example: its definition involves a choice between cases, and that choice might change from one index to the next. Trying to guess which term wins at every index can make a convergence proof unnecessarily difficult.
Instead, an algebraic identity expresses the maximum and minimum using absolute value. Earlier in this course, we proved that sums, differences, scalar multiples, and absolute values preserve limits. Combining those results gives a general theorem for maxima and minima of convergent sequences. It applies whether the same term wins every time, whether the winner changes, or whether the two sequences approach the same limit.
Two Identities That Resolve the Cases
For real numbers \(x\) and \(y\), \(\max\{x,y\}\) means the larger of the two numbers, and \(\min\{x,y\}\) means the smaller. The identities below package the two possible orderings into one formula.
To verify the identities, first suppose \(x\geq y\). Then \(|x-y|=x-y\), so $$ \frac{x+y+|x-y|}{2} =\frac{x+y+x-y}{2}=x $$ and $$ \frac{x+y-|x-y|}{2} =\frac{x+y-x+y}{2}=y. $$ These are the maximum and minimum, respectively. If \(x<y\), then \(|x-y|=y-x\), and the same expressions become $$ \frac{x+y+y-x}{2}=y, \qquad \frac{x+y-(y-x)}{2}=x. $$ Again, they are the maximum and minimum. The cases \(x\geq y\) and \(x<y\) cover every pair of real numbers, so the identities hold in all cases.
For sequences, apply these identities separately at each index. The identities do not require the ordering of the terms to remain fixed: the absolute value automatically accounts for whichever ordering holds at that index.
Limits of Maxima and Minima
Proof. By the Limit of a Difference, \(a_n-b_n\to L-M\). By the theorem that absolute values preserve limits, $$ |a_n-b_n|\longrightarrow|L-M|. $$ The Limit of a Sum and the Limit of a Scalar Multiple therefore give $$ \frac{a_n+b_n+|a_n-b_n|}{2} \longrightarrow \frac{L+M+|L-M|}{2}. $$ By the maximum identity, the sequence on the left is \(\max\{a_n,b_n\}\), and the expression on the right is \(\max\{L,M\}\). This proves the first limit.
For the minimum, the same limit laws give $$ \frac{a_n+b_n-|a_n-b_n|}{2} \longrightarrow \frac{L+M-|L-M|}{2}. $$ The minimum identity identifies the left side as \(\min\{a_n,b_n\}\) and the right side as \(\min\{L,M\}\). This proves the second limit. \(\square\)
The theorem gives the limit even if there is no fixed index after which one sequence always supplies the maximum. It uses only established limit laws and identities, so there is no need to divide the proof into infinitely many possible patterns of winners.
When the Larger Limit Eventually Wins
If the limits are different, the sequence with the larger limit eventually supplies the maximum at every index. This conclusion is stronger than knowing the limit of the maximum: it describes which term is selected for all sufficiently large indices. The proof uses the Eventual Ordering from Distinct Limits theorem established earlier in the course.
Proof. By the Eventual Ordering from Distinct Limits theorem, since \(L<M\), there is an \(N\in\mathbb{N}_0\) such that \(a_n<b_n\) for every \(n\geq N\). Whenever \(a_n<b_n\), the definition of maximum gives \(\max\{a_n,b_n\}=b_n\), and the definition of minimum gives \(\min\{a_n,b_n\}=a_n\). These equalities therefore hold for every \(n\geq N\). \(\square\)
When \(L>M\), interchange the roles of the two sequences: eventually \(b_n<a_n\), so the maximum is \(a_n\) and the minimum is \(b_n\). When \(L=M\), this argument does not give eventual selection. The sequences may change which one is larger indefinitely, while their maximum and minimum still converge to the common limit.
Worked Examples: Applying the Identities
Worked Example: A Maximum Whose Winner Changes
For \(n\in\mathbb{N}_0\), define $$ p_n=2+\frac{1}{n+1}, \qquad q_n=3-\frac{2}{n+1}. $$ The reciprocal sequence \(1/(n+1)\) tends to zero, so the limit laws give \(p_n\to2\) and \(q_n\to3\). The Limits of Maxima and Minima theorem now shows that $$ \max\{p_n,q_n\}\longrightarrow3, \qquad \min\{p_n,q_n\}\longrightarrow2. $$
The winning term really does change. Subtracting gives $$ p_n-q_n =2+\frac{1}{n+1}-3+\frac{2}{n+1} =-1+\frac{3}{n+1}. $$ At \(n=0\), this difference is \(2\), so \(p_0=3\) and \(q_0=1\). At \(n=1\), it is \(1/2\), so \(p_1=5/2\) and \(q_1=2\). At \(n=2\), it is \(0\), and both terms equal \(7/3\). For every \(n\geq3\), \(n+1\geq4\), so \(3/(n+1)\leq3/4<1\), which makes \(p_n-q_n<0\). Thus \(q_n\) eventually wins, in agreement with the distinct-limit theorem. The maximum limit theorem establishes the limits without needing this index-by-index comparison.
Worked Example: The Winner Changes Forever
Let $$ u_n=\frac{(-1)^n}{n+1}, \qquad v_n=-\frac{(-1)^n}{n+1}. $$ Since \(|u_n|=|v_n|=1/(n+1)\to0\), the Absolute-Value Criterion for Convergence to Zero gives \(u_n\to0\) and \(v_n\to0\). The maximum and minimum theorem implies $$ \max\{u_n,v_n\}\longrightarrow0, \qquad \min\{u_n,v_n\}\longrightarrow0. $$
At each index, one term is positive and the other is negative. If \(n\) is even, then \(u_n=1/(n+1)\) and \(v_n=-1/(n+1)\), so \(u_n\) is the maximum. If \(n\) is odd, then \(u_n=-1/(n+1)\) and \(v_n=1/(n+1)\), so \(v_n\) is the maximum. Consequently, neither sequence supplies the maximum eventually. The maximum and minimum nevertheless tend to zero because both candidate values approach zero.
The example shows why it is important not to assume that one sequence eventually wins when the two limits are equal. The identity-based theorem covers this situation directly.
Worked Example: Clipping Terms to an Interval
Define \(x_n=1+(-1)^n/(n+1)\), and form the clipped sequence $$ c_n=\max\{0,\min\{1,x_n\}\}. $$ Because \(|x_n-1|=1/(n+1)\to0\), we have \(x_n\to1\). The minimum theorem applied to the constant sequence \(1\) and the sequence \(x_n\) gives $$ \min\{1,x_n\}\longrightarrow\min\{1,1\}=1. $$ Applying the maximum theorem to the constant sequence \(0\) and this minimum sequence then gives \(c_n\to\max\{0,1\}=1\).
The clipping has the intended effect at every index. If \(n\) is even, then \(x_n=1+1/(n+1)>1\), so \(\min\{1,x_n\}=1\) and \(c_n=1\). If \(n\) is odd, then \(n+1\geq2\), and $$ \frac12\leq 1-\frac{1}{n+1}=x_n<1. $$ Thus \(\min\{1,x_n\}=x_n>0\), and \(c_n=x_n\). Every clipped term lies in \([0,1]\), and the theorem proves that the clipped sequence still approaches \(1\).
A Reliable Proof Strategy and a Common Pitfall
When a sequence is defined using a maximum or minimum, first identify the candidate sequences and their limits. Then apply the relevant identity or the Limits of Maxima and Minima theorem. If the limits are distinct and the actual eventual choice matters, use the Eventual Selection theorem. If the limits are equal, do not infer that the same term must win eventually; the alternating example shows otherwise.
A common error is to replace \(\max\{a_n,b_n\}\) by \(a_n\) merely because \(a_n\) appears to be larger in a few early calculations, or because its limit is larger without checking eventual ordering. A finite collection of calculations does not establish an inequality for all sufficiently large indices. When the limits are distinct, the earlier eventual-ordering result supplies the needed justification. When they are equal, the winner may keep changing, but the maximum and minimum still converge by the identities and limit laws.
These methods apply to any expression built by finitely many maxima and minima of convergent sequences: evaluate the corresponding maximum or minimum of the limits at each step. The key is to justify each operation with the theorem, rather than guessing which branch is selected at individual indices.
Check Your Understanding
Use the identities and theorems in this tutorial to answer the following questions.
- How do the absolute-value identities express the maximum and minimum of two real numbers?
- If \(a_n\to4\) and \(b_n\to-2\), what are the limits of \(\max\{a_n,b_n\}\) and \(\min\{a_n,b_n\}\)?
- Why does a strict inequality between the two limits imply that one sequence eventually supplies the maximum?
- Can the sequence supplying the maximum change infinitely often when both sequences converge to the same limit? Give a reason.
- Which limit theorem justifies applying clipping by a fixed lower and upper bound to a convergent sequence?