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

Case 1: $l=m$
Suppose $l=m$ 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 $y_n \lt x_n$ then:
 * $\size {\max \set {x_n,y_n} - \max \set {l,m}} = \size {x_n - l} \lt \epsilon$
 * if $x_n \le y_n$ then:
 * $\size {\max \set {x_n,y_n} - \max \set {l,m}} = \size {y_n - m} \lt \epsilon$

In either case:
 * $\size {\max \set {x_n,y_n} - \max \set {l,m}} \lt \epsilon$

The result follows.

Case 2: $l > m$
Suppose $l > m$ 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:
 * $n > N_1$
 * $n > N_2$

By Corollary 3 of Negative of Absolute Value then $\forall n > N$:
 * $\size {x_n - l} < \epsilon' \implies x_n \gt l - \epsilon'$
 * $\size {y_n - m} < \epsilon' \implies y_n \lt m + \epsilon'$

Since $\epsilon' \le \delta$ then

and

So:
 * $x_n \gt l-\epsilon' \ge \dfrac {l + m} 2 \ge m + \epsilon' \gt y_n$.

Hence $x_n = \max \set{x_n,y_n}$.

So $\forall n > N$:
 * $\size {\max \set {x_n,y_n} - \max \set {l,m}} = \size {x_n - l} \lt \epsilon' \le \epsilon$

The result follows.

Case 3: $m > l$
Similar to Case 2 by interchanging $l$ with $m$ and $x_n$ with $y_n$.

Also see

 * Minimum Rule for Real Sequences