Sequence Converges to Within Half Limit

Sequence of Real Numbers
Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$ or $\Q$.

Let $\left \langle {x_n} \right \rangle$ be convergent to the limit $l$.

That is, let $\displaystyle \lim_{n \to \infty} x_n = l$.

Suppose $l > 0$.

Then $\exists N: \forall n > N: x_n > \dfrac l 2$.

Similarly, suppose $l < 0$.

Then $\exists N: \forall n > N: x_n < \dfrac l 2$.

Sequence of Complex Numbers
Let $\left \langle {z_n} \right \rangle$ be a sequence in $\C$.

Let $\left \langle {z_n} \right \rangle$ be convergent to the limit $l$.

That is, let $\displaystyle\lim_{n \to \infty} z_n = l$ where $l \ne 0$.

Then:
 * $\displaystyle\exists N: \forall n > N: \left\vert{z_n}\right\vert > \frac {\left\vert{l}\right\vert} 2$

Proof for Sequence of Real Numbers
Suppose $l > 0$.

From the definition of convergence to a limit:
 * $\forall \epsilon > 0: \exists N: \forall n > N: \left\vert{x_n - l}\right\vert < \epsilon$

That is, $l - \epsilon < x_n < l + \epsilon$.

As this is true for all $\epsilon > 0$, it is also true for $\epsilon = \dfrac l 2$ for some value of $N$.

Thus:
 * $\exists N: \forall n > N: x_n > \dfrac l 2$

as required.

Now suppose $l < 0$.

By a similar argument:
 * $\forall \epsilon > 0: \exists N: \forall n > N: l - \epsilon < x_n < l + \epsilon$

Thus it is also true for $\epsilon = \dfrac l 2$ for some value of $N$.

Thus:
 * $\exists N: \forall n > N: x_n < \dfrac l 2$

as required.

Proof for Sequence of Complex Numbers
Suppose $l > 0$.

Let us choose $N$ such that $\displaystyle \forall n > N: \left\vert{z_n - l}\right\vert < \frac {\left\vert{l}\right\vert} 2$.

Then:

Note
Although this result seems a little trivial, it is often crucial to know that a sequence will be "eventually non-zero" so we know we can legitimately divide by it.

This is used in the Quotient Rule in Combination Theorem for Sequences.