What would be the easiest way to prove that the lim (as n approaches infinity) of (2)/(n+1)=0.

Thanks.

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- September 28th 2006, 05:06 PMJaysFan31Limit Proof
What would be the easiest way to prove that the lim (as n approaches infinity) of (2)/(n+1)=0.

Thanks. - September 28th 2006, 05:14 PMPlato
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. - September 28th 2006, 05:46 PMThePerfectHacker
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