In $R$, given that $\lim_{n\to\infty} {x_n}= L,$ prove that $\lim_{n\to\infty} \frac{1}{x_n}=\frac{1}{L}.$
Here is my attempt at the proof.
"We need an $N$: $n>N\implies\left|\frac{1}{x_n}-\frac{1}{L}\right|<\epsilon.$
$\left|\frac{1}{x_n}-\frac{1}{L}\right| =\frac{|x_n-L|}{|x_n||L|}$
Since $x_n$ is convergent, it is bounded and $\exists$$N_1$: n>$N_1$⟹ |$x_n$|>$\frac{|L|}{2}$
$\implies\frac{1}{|x_n|}<\frac{2}{|L|}$
and since $\lim_{n\to\infty} {x_n}=L,$ $\exists N_2$: n>N_2\implies|x_n$-L|<\frac{\epsilon|L|^2}{2}.$
Now, when $n>\max\{N_1,N_2\},$
(|$x_n$-L|)($\frac{1}{|x_n|}$) <($\frac{\epsilon|L|^2}{2}$)($\frac{2}{|L|}$) = $\epsilon|L|$
and then finish the proof my dividing by |L|..."
I'm really new to this style of proof so any nick picky advice or corrections are welcome!