Sequence Converges to Within Half Limit/Real Numbers

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Theorem

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

Let $\sequence {x_n}$ be convergent to the limit $l$.

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


Suppose $l > 0$.

Then:

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


Similarly, suppose $l < 0$.

Then:

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


Proof

Suppose $l > 0$.

From the definition of convergence to a limit:

$\forall \epsilon > 0: \exists N: \forall n > N: \size {x_n - l} < \epsilon$

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.

$\blacksquare$


Sources