# Another limit proof

• Oct 10th 2008, 11:54 PM
hayter221
Another limit proof
Prove if lim(x sub n)=x and x>0 then there exists a natural number M such that x sub n > 0 for all n> or equal to M

Any hints on how to start this or what the end result should be would be greatly appreciated, thanks.
• Oct 11th 2008, 12:01 AM
CaptainBlack
Quote:

Originally Posted by hayter221
Prove if lim(x sub n)=x and x>0 then there exists a natural number M such that x sub n > 0 for all n> or equal to M

Any hints on how to start this or what the end result should be would be greatly appreciated, thanks.

if ${\rm{lim}}_{n \to \infty}{x_n}=x,\ \ x>0$ then for all $\varepsilon>0$ there exists an $N_{\varepsilon}$ such that:

$|x_n-x|<\varepsilon$ for all $n\ge N_{\varepsilon}$

Now choose $\varepsilon=x/2$ and you should find the result follows with $M=N_{\varepsilon}.$

CB
• Oct 11th 2008, 12:08 AM
hayter221
I'm not sure I follow, what do I do with the epsilon...sorry I have just begun to learn this and I haven't fully grasped the idea of the limit definition.
• Oct 11th 2008, 05:45 AM
CaptainBlack
Quote:

Originally Posted by hayter221
I'm not sure I follow, what do I do with the epsilon...sorry I have just begun to learn this and I haven't fully grasped the idea of the limit definition.

You can choose anything you like for $\varepsilon$, that is the definition of convergence for a sequence.

What we do is choose a particular value to ensure that $x_n$ is positive from $N_{\varepsilon}$ onwards. We do this by choosing any value of $\varepsilon \le x/2$ then as $|x_n-x|<\varepsilon$ for all $n \ge N_{\varepsilon}$ we have:

$|x_n-x|

so:

$-x/2

rearranging:

$x/2

for all $n \ge N_{\varepsilon}$

But $x>0$, so:

$x_n>0$ for all $n \ge N_{\varepsilon}$

CB
• Oct 11th 2008, 11:38 AM
hayter221
thanks, much appreciated!