# Limit Proof

• Oct 9th 2007, 09:06 AM
jlindbo
Limit Proof
I am having a bit of trouble with what should be a simple proof:

Let (an) be a convergent sequence of real numers, an >= 0 for all n in the natural number set. If the limit of an is a (an->a) show that a >= 0.

I understand that I need to use an epsilon neighbourhood to show a contradiction when a is < 0 but am having a hard time finding an epislon that produces a contradiction.

Thanks for the help
• Oct 9th 2007, 09:19 AM
Plato
Suppose that a<0 then choose $\displaystyle \varepsilon = \frac{{ - a}}{2} > 0$.
Because of sequence convergence we have $\displaystyle \left( {\exists N} \right)\left[ {n \ge N \Rightarrow \left| {a_n - a} \right| < \varepsilon } \right].$

But that means: $\displaystyle a - \varepsilon < a_n < a + \varepsilon \le a + \frac{{ - a}}{2} = \frac{a}{2} < 0$.

• Oct 9th 2007, 09:46 AM
jlindbo
$\displaystyle \le a + \frac{{ - a}}{2} = \frac{a}{2} < 0$

I am sure I am just being very dense but where did the rhs of the inequality come from?
• Oct 9th 2007, 10:05 AM
Jhevon
Quote:

Originally Posted by jlindbo
$\displaystyle \le a + \frac{{ - a}}{2} = \frac{a}{2} < 0$

I am sure I am just being very dense but where did the rhs of the inequality come from?

$\displaystyle \epsilon = - \frac a2$, so $\displaystyle a + \epsilon = a - \frac a2 = \frac a2$

since $\displaystyle - \frac a2 > 0$, we have $\displaystyle \frac a2 < 0$

now continue to show the contradiction
• Oct 9th 2007, 11:19 AM
jlindbo
Doh ;P

And since $\displaystyle a_n > 0$ that produces the contradiction.

Thanks
• Oct 9th 2007, 11:22 AM
Jhevon
Quote:

Originally Posted by jlindbo
Doh ;P

And since $\displaystyle a_n > 0$ that produces the contradiction.

Thanks

correct
• Oct 9th 2007, 12:28 PM
Plato
This is a hugely important idea in sequence convergence that has a similar application to continuous functions.

If the limit of a sequence is positive then almost all the terms of that sequence must be positive. There is a similar statement of a negative limit.