By inspection, it's clear that the sequence must converge to 
because
when
is arbitrarily large.
Now, for the limit as
to be equal to is to say that for any 
, there exists some 
such that whenever 
, it follows that
From this inequality, we get
As we're considering, we can omit the first inequality.
We can then see that choosing
will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that 
.
when
Now, for the limit as
From this inequality, we get
As we're considering
We can then see that choosing