# 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:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l$
$\displaystyle \lim_{n \mathop \to \infty} y_n = m$

Then:

$\displaystyle \lim_{n \mathop \to \infty} \max \set {x_n, y_n} = \max \set {l, m}$

## 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 $\epsilon' \le \delta$:

 $\displaystyle l - \epsilon'$ $\ge$ $\displaystyle l - \delta$ $\displaystyle$ $=$ $\displaystyle l - \dfrac {l - m} 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {2 l - l + m} 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {l + m} 2$

and:

 $\displaystyle m + \epsilon'$ $\le$ $\displaystyle m + \delta$ $\displaystyle$ $=$ $\displaystyle m + \dfrac {l - m} 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {2 m + l - m} 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {l + m} 2$

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 $y_n$.

$\blacksquare$