Solving Supremum Question: Is 4 the Right Answer?

[Cankur]

Hello!

I had a test in which the question that I will present here was asked. I got no points for my attempt at a solution. Do you think that I was still on the right track and that I deserve partial points? Here is the question:

"A number M is said to be an upper bound to a set A if M \geq x for every x\in A. A number S is said to be supremum of a set A if S is the smallest upper bound to A.

Assume that:

A = {(4n²)/(n²+1) : n \geq0 is an integer}.

Show that supremum of A is 4."

And here is what I wrote as an answer (not verbatim, but translated from another language):

"Since n does not have an upper limit, it can go toward infinity. In this case:

A = lim (n \rightarrow \infty) (4n²)/(n²+1)=\infty/\infty. This shows that we can use l'hopital's rule. After using l'hopital's rule twice we get that A = 4. In other words, this gives us supremum. Since n always can be even bigger, this is just the smallest upper bound.

Answer: By using l'hopital's rule twice, I have shown that supremum A is 4."

Out of the possible 4 points that one could get on that question, I got 0. Was it justified?

Thanks in advance!

 

[HallsofIvy]

Essentially, what you showed was that the limit, as n goes to infinity, of that sequence is 4. That does NOT prove that 4 is the supremum. For example, the limit of 1/n, as n goes to infinity is 0 but 0 is definitely not the supremum!

Here, you would also have to show that your sequence is increasing and you did not do that.

Oh, and I certainly would not have used L'Hopital's rule for that limit: just divide both numerator and denominator by n².

 

[mathman]

In simple terms: (4n²)/(n²+1)=4/(1+1/n²) < 4.

As n becomes infinite limit is 4. Another way is assume the sup = a < 4, then you can find a large enough n so the expression > a: contradiction.

 

[Cankur]

Thanks for the answers! I see that there were quite essential things that I missed. But you don't think I deserve some points for my answer?