Lower and Upper Bounds for Sequences

Theorem

Let $\sequence {x_n}$ be a sequence in $\R$.

Let $x_n \to l$ as $n \to \infty$.

Then:

$(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$
$(2): \quad \forall n \in \N: x_n \le b \implies l \le b$

Corollary

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:

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

Let there exist $N \in \N$ such that:

$\forall n \ge N: x_n \le y_n$

Then:

$l \le m$

Proof

$(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$:

Let $\epsilon > 0$.

Then:

$\exists N \in \N: n > N \implies \size {x_n - l} < \epsilon$

So from Negative of Absolute Value:

$l - \epsilon < x_n < l + \epsilon$

But $x_n \ge a$, so:

$a \le x_n < l + \epsilon$

Thus, for any $\epsilon > 0$:

$a < l + \epsilon$

From Real Plus Epsilon it follows that $a \le l$.

$\Box$

$(2): \quad \forall n \in \N: x_n \le b \implies l \le b$:

If $x_n \le b$ it follows that $-x_n \ge -b$ and the above result can be used.

$\blacksquare$

Warning

Let $\sequence {x_n}$ be a sequence in $\R$.

Let $x_n \to l$ as $n \to \infty$.

Then it is not necessarily the case that:

$(1): \quad \forall n \in \N: x_n > a \implies l > a$
$(2): \quad \forall n \in \N: x_n < b \implies l < b$