Can someone check my process of proving "if lim(f/g)=0, then f(n)=o(g(n))"?

Question

I am trying to prove the following statement.

if $\displaystyle \lim_{n\rightarrow\infty}\frac{f(n)}{g(n)}=0$, then $f(n)$ is $o(g(n))$

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My solution

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$o\left(g(n)\right)=\{f(n):\forall c>0, \exists n_0, \forall n>n_0, 0 \le f(n) \lt cg(n)\}$

$0 \le f(n) \lt cg(n)$ $\Leftrightarrow$ $0 \le \displaystyle\frac{f(n)}{g(n)} \lt c$ $\Leftrightarrow$ $\displaystyle \lim_{n\rightarrow\infty}0 \le \lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} \le \lim_{n\rightarrow\infty}c$

$\displaystyle\Leftrightarrow$ $0 \le \displaystyle\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} \le c$

$\displaystyle\Leftrightarrow$ $0 \le \displaystyle\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} \le 0$ ($\because$ c is any positive constant number)

$\displaystyle\Leftrightarrow$ $\displaystyle\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = 0$

$\therefore$ if $\displaystyle \lim_{n\rightarrow\infty}\frac{f(n)}{g(n)}=0$, then $f(n)$ is $o(g(n))$.

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Is my solution right mathematically?

I was wondering if there exists any holes in my process.

(e.g. omitting any conditions)

Thank you for reading my question.

Answer 1

By definition $\lim_{n\to+\infty}a_n=A$ means that $$ \forall\epsilon>0\ \exists n_0\in\mathbb{N}\ \forall n\ge n_0\colon |a_n-A|\le\epsilon. $$

Apply it to $\lim_{n\to+\infty}\frac{f(n)}{g(n)}=0$ $$ \forall\epsilon>0\ \exists n_0\in\mathbb{N}\ \forall n\ge n_0\colon \underbrace{\left|\frac{f(n)}{g(n)}\right|\le\epsilon}_{\Leftrightarrow\ |f(n)|\le\epsilon|g(n)|} $$ and compare what you've got with the definition of the Little-O notation to conclude that $$ f(n)=o(g(n)), \ n\to+\infty. $$