1. ## 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.

2. Originally Posted by hayter221
Prove lim(x sub n) = 0 if and only if lim(|x sub n|)= 0.
Surely you know this: $\left| {\left| {x_n } \right| - 0} \right| = \left| {\left| {x_n } \right|} \right| = \left| {x_n } \right| = \left| {x_n - 0} \right|$

3. 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.

4. well, by assumption, $\lim x_n = 0$ iff for every $\varepsilon>0$ there exists a natural number $K$ such that for all $n \geq K$ we have $|x_n - 0| < \varpesilon$..

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

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