If lim s_n =0 that means,

|s_n|<e for n>N

Hence if you chose n>N we have,

||s_n|-0|=||s_n||=|s_n|<e

Thus,

lim |s_n| = 0

The same the other way around.

Let lim s_n = L2) Let (Sn) be a convergent sequence, and suppose that lim Sn > a.

Prove that htere exists a number N such that n > N implies Sn > a.

.

By hypothesis L>a thus, L-a>0 thus (L-a)/2 >0

That means,

|s_n-L| < (L-a)/2 for n>N

Open absolute values,

-(L-a)/2< s_n - L < (L-a)/2

Thus,

L-(L-a)/2 < s_n < L + (L-a)/2

Thus,

(L+a)/2 < s_n < (2L-a)/2

The fist inequality says,

(L+a)/2 < s_n

Thus,

(a+a)/2 < (L+a)/2 < s_n for n>N

Thus,

a < s_n for all n>N