Maximum Rule for Real Sequences
Theorem
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
- $\ds \lim_{n \mathop \to \infty} x_n = l$
- $\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
- $\ds \lim_{n \mathop \to \infty} \max \set {x_n, y_n} = \max \set {l, m}$
Proof
Case $1$: $l = m$
Let $l = m$.
Then:
- $\max \set {l, m} = l = m$
Let $\epsilon > 0$ be given.
By definition of the limit of a real sequence, we can find $N_1$ such that:
- $\forall n > N_1: \size {x_n - l} < \epsilon$
Similarly we can find $N_2$ such that:
- $\forall n > N_2: \size {y_n - m} < \epsilon$
Let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true:
- $n > N_1$
- $n > N_2$
Thus $\forall n > N$:
- if $y_n < x_n$ then:
- $\size {\max \set {x_n, y_n} - \max \set {l, m} } = \size {x_n - l} < \epsilon$
- if $x_n \le y_n$ then:
- $\size {\max \set {x_n, y_n} - \max \set {l, m} } = \size {y_n - m} < \epsilon$
In either case:
- $\size {\max \set {x_n, y_n} - \max \set {l, m} } < \epsilon$
The result follows.
$\Box$
Case 2: $l > m$
Let $l > m$.
Then:
- $l = \max \set {l, m}$
Let $\delta = \dfrac {l - m} 2$.
Then:
- $\delta > 0$
Let $\epsilon > 0$ be given.
Let $\epsilon' = \min \set {\delta, \epsilon}$.
Then:
- $\epsilon' > 0$
By definition of the limit of a real sequence, we can find $N_1$ such that:
- $\forall n > N_1: \size {x_n - l} < \epsilon'$
Similarly we can find $N_2$ such that:
- $\forall n > N_2: \size {y_n - m} < \epsilon'$
Let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true:
- $n > N_1$
- $n > N_2$
From Corollary 3 of Negative of Absolute Value, we have that $\forall n > N$:
- $\size {x_n - l} < \epsilon' \implies x_n > l - \epsilon'$
- $\size {y_n - m} < \epsilon' \implies y_n < m + \epsilon'$
As $\epsilon' \le \delta$:
| \(\ds l - \epsilon'\) | \(\ge\) | \(\ds l - \delta\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds l - \dfrac {l - m} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {2 l - l + m} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {l + m} 2\) |
and:
| \(\ds m + \epsilon'\) | \(\le\) | \(\ds m + \delta\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds m + \dfrac {l - m} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {2 m + l - m} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {l + m} 2\) |
So:
- $x_n > l - \epsilon' \ge \dfrac {l + m} 2 \ge m + \epsilon' > y_n$
Hence:
- $x_n = \max \set{x_n, y_n}$
So $\forall n > N$:
- $\size {\max \set {x_n, y_n} - \max \set {l, m} } = \size {x_n - l} < \epsilon' \le \epsilon$
The result follows.
$\Box$
Case 3: $m > l$
Similar to Case 2 by interchanging $l$ with $m$ and $x_n$ with $y_n$.
$\blacksquare$