# Limit Proof

• Oct 10th 2008, 02:46 PM
hayter221
Limit Proof
Prove lim(x sub n) = 0 if and only if lim(|x sub n|)= 0.

I am not sure how to even start this or what property of limits apply here.
• Oct 10th 2008, 03:06 PM
Plato
Quote:

Originally Posted by hayter221
Prove lim(x sub n) = 0 if and only if lim(|x sub n|)= 0.

Surely you know this: $\displaystyle \left| {\left| {x_n } \right| - 0} \right| = \left| {\left| {x_n } \right|} \right| = \left| {x_n } \right| = \left| {x_n - 0} \right|$
• Oct 10th 2008, 06:40 PM
hayter221
Can you show me how to incorporate this into the proof? I have that for every epsilon>0 there exists a natural number K(epsilon) such that for all n>or equal to K(epsilon) the tems x sub n satisfy |x sub n -0|< epsilon. From what you said |x sub n| = |x sub n -0| but I am not sure how this leads to lim(|x sub n|) = 0 from lim(x sub n)=0 or vice versa.
• Oct 11th 2008, 05:22 AM
kalagota
well, by assumption, $\displaystyle \lim x_n = 0$ iff for every $\displaystyle \varepsilon>0$ there exists a natural number $\displaystyle K$ such that for all $\displaystyle n \geq K$ we have $\displaystyle |x_n - 0| < \varpesilon$..

and from plato's post, the last in the equality is less than epsilon..

in the same manner, $\displaystyle \lim |x_n| = 0$ iff for every $\displaystyle \varepsilon>0$ there exists a natural number $\displaystyle K$ such that for all $\displaystyle n \geq K$ we have $\displaystyle ||x_n| - 0| < \varpesilon$..