Proof of limit of quotient sequence property I am struggling to prove the following property of limits of sequences:

If $(a_n)_{n\in\mathbb{N}}$ is a sequence such that $\forall n:\ a_n\ne0$ and $\lim\limits_{n\to\infty} a_n = L \ne 0$, then
  $$\lim\limits_{n\to\infty} \frac{1}{a_n} = \frac{1}{L}$$

Using the definition of limit,
$$\forall \varepsilon>0\ \exists n_0\ \forall n \ge n_0\quad \left|a_n - L\right| < \varepsilon$$
what I have up to now is
$$\begin{equation}\label{eq:inv-limit}
    \left|{\frac{1}{a_n}-\frac{1}{L}}\right| = 
    \left|{\frac{L-a_n}{a_nL}}\right| = 
    \frac{\left|a_n- L\right|}{\left|a_nL\right|} < 
    \frac{\varepsilon}{\left|a_n\right|\left|L\right|}\,.
   \end{equation}$$
But now the coefficient of epsilon is not constant, so that is not sufficient to prove that $\frac{1}{a_n}\to\frac{1}{L}$. How to proceed?
 A: You are absolutely right. At the final step notice the that $|a_n|$ can be bounded from below and above since $$a_n\to L$$and we have$$0<L-\epsilon'<a_n<L+\epsilon'$$where $L-\epsilon'$ can be arbitrarily a large positive number since $L$ is positive and $\epsilon'>0$ is arbitrary
A: You are close.
You have
$\frac{1}{a_n}-\frac{1}{L}
 = \frac{\left|a_n- L\right|}{\left|a_nL\right|}
$.
Since
$L \ne 0$
and $a_n \to L$,
for any $c > 0$
there is an $n(c)$ such that
$|a_n-L| < c$
for $n > n(c)$.
Choosing $c = |L|/2$,
then
$|a_n-L| < |L|/2$
for $n > n(|L|/2)$.
Therefore
$|a_n| > |L|/2$
for such $n$
so that
$|a_nL| > |L|^2/2$
For this $c$,
$\frac{1}{a_n}-\frac{1}{L}
 = \frac{\left|a_n- L\right|}{\left|a_nL\right|}
\lt \frac{\left|a_n- L\right|}{\left|L^2/2\right|}
$.
To make
$|\frac{1}{a_n}-\frac{1}{L}|
\lt \epsilon$,
it is enough to make
$\frac{\left|a_n- L\right|}{\left|L^2/2\right|}
\lt \epsilon$,
or
$\left|a_n- L\right|
\lt \left|L^2/2\right|\epsilon
$.
Combining these two bounds
if
$n > max(n(|L|/2), n( \left|L^2/2\right|\epsilon))$,
then
$|\frac{1}{a_n}-\frac{1}{L}|
\lt \epsilon$.
