Alex M.
asked 07/14/20Proving delta epsilon limits at infinity for unknown function
I have a question and it's limx>inf (f(x)= 2 and I have to prove that N exists and if x>N 1<f(x)<3
I started trying to set up my proof with |f(x)-2|<E but I've never done a problem like this with an unknown function and I have no idea what to do next. If anyone can help solve this or let me know where to start, it would be greatly appreciated!
1 Expert Answer
Sava D. answered 07/14/20
Math master degree tutor with good understanding of calculus
Here is one way to word this problem.
We are given that lim (x->∞) (f(x)) = 2. This means that for any positive number ε, there exists a number Nε, such that when
x > Nε,
then
|f(x) - 2| < ε.
Since the above statement is true, we know, that there is a positive number N1, such that
when
x > N1
then
|f(x) - 2| < 1.
With other words,
-1 < f(x) - 2 < 1,
1 < f(x) < 3.
Sava D.
07/14/20
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Stanton D.
Hi Alex M., It would appear that you are about there. The idea of lim x->.infinity. implies exists n such that |f(x)-2|< epsilon. Therefore, setting epsilon to 1 is just a special instance of that. That doesn't GIVE you the least possible N, of course, since you don't have the function specified! -- Cheers, --Mr. d.07/14/20