In computing the asymptotic complexity of the sum of two functions, one theorem states that if $\large\lim_{n\rightarrow\infty}\frac{f_2(n)}{f_1(n)}$ exists, then the asymptotic complexity is $f_1(n)$. The author gave an example of adding $f_1(n)=n^3$ and $f_2(n)=n^2$. He determined that the limit exists and concluded that the complexity was $O(f_1(n))$ or $O(n^3)$.
Obviously the value of the functions matter because if we switched it around and made $f_1(n)=n^2$ and $f_2(n)=n^3$, then we couldn't compute the complexity that way as $\large\lim_{n\rightarrow\infty}\frac{n^3}{n^2} = \infty$ which does not exist.
Is this correct? If so, how do I make sure I am adding the right way?
Your calculations are correct, but I think it's the phrasing that's confused, order doesn't matter here in quite the way you seem to think it does. (If I'm interpreting your statement correctly.) Carefully phrased, you've found:
Because $$\lim \frac{n^2}{n^3} = 0$$ we can say that $n^2$ is $O(n^3)$ (it's also $O(n^2)$). Because $$\lim\frac{n^3}{n^2} =\infty$$ we can say that $n^3$ is not $O(n^2)$.
I think maybe the example you were looking for is:
Since $$\lim\frac{n^3+n^2}{n^3} =1$$ we know that $n^3 +n^2 = n^2 +n^3$ is $O(n^3)$.
A major use of this (in a slightly different context $x\to 0$) is to make statements of the form: $$e^x = 1 + x + O(x^2)$$ Hope it helps.