Lower and Upper Bounds for Sequences

Theorem
Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

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

Then:


 * $\forall n \in \N: x_n \ge a \implies l \ge a$;
 * $\forall n \in \N: x_n \le b \implies l \le b$.

Proof

 * $\forall n \in \N: x_n \ge a \implies l \ge a$:

Let $\epsilon > 0$.

Then $\exists N \in \N: n > N \implies \left|{x_n - l}\right| < \epsilon$.

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 $a \le l$.


 * $\forall n \in \N: x_n \le b \implies l \le b$:

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

Warning
Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

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

Then it is not the case that:


 * $\forall n \in \N: x_n > a \implies l > a$;
 * $\forall n \in \N: x_n < b \implies l < b$.

Take the examples:


 * $\displaystyle \left \langle {x_n} \right \rangle = \frac 1 n$
 * $\displaystyle \left \langle {y_n} \right \rangle = -\frac 1 n$

Then :
 * $\displaystyle \forall n \in \N^*: \frac 1 n > 0, -\frac 1 n < 0$.

From Power of Reciprocal: Corollary, we have
 * $x_n \to 0$
 * $y_n \to 0$

as $n \to \infty$.

However, it is clearly false that $0 > 0$ and $0 < 0$.