This one is easy: 2/[n+1] < 2/n.
If e>0 then there is a positive integer K such that (1/K)<(e/2) or (2/K)<e.
Note the sequnce,
a_n={2/(n+1)}
Is identical to the sequence,
b_n={2(1/n)/(1+1/n)}
Now the numerator is,
lim 2(1/n)
Since, lim (1/n)--->0
So too, lim 2(1/n)---0 by constant function rule.
Next, the demonator is,
lim (1+1/n)
But, lim 1=1 and lim (1/n)=0
Thus by the rule of seuqnces addition,
lim (1+1/n) exists and is, 1+0=1 (not equal to zero)
Thus, by the rule of sequnces division,
lim 2(1/n)/(1+1/n)--->0
Thus,
lim a_n--->0