Prove using the #epsilon-delta# method that #lim_(n->oo)= (2n)/(n^2+1) = 0# ?

1 Answer
Jan 28, 2018

We have to prove that given #epsilon>0# there exists #M# such that if #n>M# then #(2n)/(n^2+1)< epsilon#

Explanation:

Although it is not clear in the statement, I am assuming here that #n# are natural numbers.

Since #n^2+1 > n^2#, we can say

#(2n)/(n^2+1)< (2n)/(n^2)=2/n #

Then let's consider any #epsilon >0#, and

#2/n < epsilon#, or equivalently #2/epsilon < n#

Now, given #epsilon >0# take any #M >2/epsilon#. Then if #n>M# we know that #n >2/epsilon#, that is #2/n < epsilon#, and so:

#(2n)/(n^2+1)< (2n)/(n^2)=2/n < epsilon# for all #n>M#