I need to prove that $n^2$ is not $o(n^2+10^{10}n)$. I thought of the limit test: $$ \lim_{n \to \infty} \frac{n^2}{n^2+10^{10}n} = 1 \Rightarrow n^2 = \Theta(n^2+10^{10}n) $$
However I'm not sure if the result of the test rules out the possibility of $o(n^2+10^{10}n)$.
One definition $f(n)=o(g(n))$ is that $\lim_{n\to\infty} f(n)/g(n)=0$. If this is the definition you're using, then showing that the limit is $1$ already shows that $f(n)\neq o(g(n))$.
The other definition is that, for every $c>0$, there is an $n_0$ such that $f(n)\leq cg(n)$ for all $n\geq n_0$. The fact that $\lim_{n\to\infty} f(n)/g(n)=1$ means that, for all $\varepsilon>0$, there is some $n_0$ such that $f(n)/g(n)>1-\varepsilon$ for all $n\geq n_0$ (this is part of the definition of "limit"). So, for all $\varepsilon>0$, we have $f(n)>(1-\varepsilon)g(n)$ for all large enough $n$. This means that, in particular, we do not have $f(n)\leq cg(n)$ for $c=1-\epsilon$, so $f(n)\neq o(g(n))$ by the alternative definition.
Without any limits: $(n^2 + 10^{10}n) / 2 ≤ n^2 ≤ n^2 + 10^{10}n$ whenever $n ≥ 10^{10}$. So for c < 1/2, we don't have $n^2 < c(n^2 + 10^{10}n)$ for all large n. Actually, not for any large n.
And $f(n) = \Theta (g(n))$ does indeed rule out that $f(n) = o (g(n))$, but is not a necessary condition.
