So we are put at a disadvantage. We have no idea what theorems and/or axioms/definitions you have to work with.

This is a fairly standard result but it does come in the sequence of well developed set of theorems.

Here is an example.

If $q>1$ then $q=(1+x)$ so

\(\displaystyle \begin{align*}q^n=(1+x)^n&\ge\dfrac{n(n-1)x^2}{2} \\\dfrac{1}{q^n}&\le\dfrac{2}{n(n-1)x^2}\\\dfrac{n}{q^n}&\le\dfrac{2}{(n-1)x^2} \end{align*}\)

Hey thank you.

We have reached to the Cantor's lemma.

A few questions on your solution:

How are we sure that n is not equal 1?

How do we know from this inequality now that the limit of the sequence n/q^n is 0? According to what?

We know that 2 is a constant sequence so its limit is 2, we know that x^2 is also constant so its limit is x^2, and (n-1) tends to infinity (Do I have to prove it somehow?) - So we have 2 at the nominator and infinity at the denominator - which theorem tells me that yields 0?

And we need to pose another 0 left to n/q^n to use the sandwich, but how do we do that?

Thank you!