Question

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

Answer 1

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`