How to prove that $n^2$ is not $o(n^2+10^{10}n)$?
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&nbsp;$1$ already shows that $f(n)\neq o(g(n))$.

The other definition is that, for every $c&gt;0$, 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 $\varepsilon&gt;0$, there is some $n_0$ such that $f(n)/g(n)&gt;1-\varepsilon$ for all $n\geq n_0$ (this is part of the definition of "limit").  So, for all&nbsp;$\varepsilon&gt;0$, we have $f(n)&gt;(1-\varepsilon)g(n)$ for all large enough&nbsp;$n$. 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.