How Do You Prove the Limit of (2n+1)/(n+1) as n Approaches Infinity?

[Questions999]

Prove that the limit when x--> infinite of (2n+1)/(n+1) =2
So for ε > 0,exists N>0 so that n>N => |x -a|< ε
What I do is | (2n+1)/(n+1) |< ε, I do the math actions and I have |-1/(n+1)| < ε... NOW,what I don't get,when I remove the absolute value,do I get 1/(n+1)<ε or NOT?

 

[STEMucator]

I believe you want to use this definition :

$\forall \epsilon>0, \exists N | n>N \Rightarrow |a_n - L| < \epsilon$

So what is $|a_n - L|$? Plug in your info and start massaging it into a suitable expression.

 

[Questions999]

I know what to do,I just want to know if the part when I remove the absolute value is correct :)

 

[Mark44]

Prove that the limit when x--> infinite of (2n+1)/(n+1) =2
So for ε > 0,exists N>0 so that n>N => |x -a|< ε
What I do is | (2n+1)/(n+1) |< ε,

This is incorrect. You should start with
|(2n + 1)/(n + 1) - 2| < ε

It looks like the above is what you were working with, but didn't write it correctly.

I do the math actions and I have |-1/(n+1)| < ε... NOW,what I don't get,when I remove the absolute value,do I get 1/(n+1)<ε or NOT?

Yes.

I know what to do,I just want to know if the part when I remove the absolute value is correct :)