Sequence Converges to Within Half Limit

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

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

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

Suppose $$l > 0$$.

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

Similarly, suppose $$l < 0$$.

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

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 $$\lim_{n \to \infty} z_n = l$$ where $$l \ne 0$$.

Then $$\exists N: \forall n > N: \left|{z_n}\right| > \frac {\left|{l}\right|} 2$$.

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

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

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

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

Thus $$\exists N: \forall n > N: x_n > \frac l 2$$ 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 $$\epsilon = \frac l 2$$ for some value of $$N$$.

Thus $$\exists N: \forall n > N: x_n < \frac l 2$$ as required.

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

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

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.