Sequence Limit Reciprocal Law Proof I am aware that this has a duplicate but I am trying to prove it differently than others.
Proposition: If $\lim_{n\to\infty} a_n = L \neq 0,$ then $\lim_{n\to\infty} \frac{1}{a_n} = \frac{1}{L}$
Proof: 
$\lim_{n\to\infty} a_n = L$, so $\exists N_2 $st $\forall n > N_2 |a_n-L|< \epsilon$
Take N = max$(N_1,N_2); $ so $\forall n>N, |a_n-L| < \epsilon  $
and $\forall n>N_1, |a_n| < \frac{1}{|L|}$ (by Lemma)
So $|\frac{1}{a_n}-\frac{1}{L}| = \frac{|a_n-L|}{|a_n|*|L|} = |a_n-L|*\frac{1}{L}*\frac{1}{a_n}$
$<\epsilon *\frac{1}{L}*L$
$<\epsilon$
I am just having difficulty proving the Lemma, which should be $|a_n|<\frac{1}{|L|}$
Any advice?
 A: Your proof attempt has a few issues. For one, it changes the limit from $A$ to $L$ (I will use $L$ here). Also, it doesn't define what $N_1$ is, but from the context it seems to be for the minimum value where $\left|a_n - \frac{1}{L}\right| \lt \epsilon$. Finally, and most importantly, you can't prove your suggested lemma of $\left| a_n \right| \lt \frac{1}{\left| L \right|}$ because it's not necessarily true. For example, if $a_n = 2 + \frac{1}{n}$, then $L = 2$, but you never have $\left|a_n\right| \lt \frac{1}{2}$.
The way I would do it instead is to expand the part in absolute values and then use the resulting inequalities to place an appropriate upper limit on $\frac{1}{\left| a_n \right|}$. In particular, you have
$$\left| a_n - L \right| \lt \epsilon \; \Rightarrow \; -\epsilon \lt a_n - L \lt \epsilon \; \Rightarrow \; -\epsilon + L \lt a_n \lt \epsilon + L \tag{1}\label{eq1}$$
There are $2$ main cases to consider, i.e., $L$ being positive or negative. I'll show how to handle the positive case. First, have $\epsilon \lt \frac{L}{2}$. As such, all $3$ parts of the end of \eqref{eq1} are positive. Thus, multiplying the left & middle parts by $\frac{1}{a_n\left(L - \epsilon\right)}$ gives
$$\frac{1}{a_n} \lt \frac{1}{L - \epsilon} \lt \frac{2}{L} \tag{2}\label{eq2}$$
with the last inequality coming from $L - \epsilon \gt \frac{L}{2}$ by the earlier assumption. This gives that
$$\left|a_n - L\right| \times \frac{1}{L} \times \frac{1}{\left|a_n\right|} \lt \frac{2\epsilon}{L^2} \tag{3}\label{eq3}$$
The epsilon to choose for proving the limit of $\frac{1}{a_n}$, call it $\epsilon_2$, could be any $\epsilon_2 = \frac{2\epsilon}{L^2} \lt L$. In that case, the original epsilon would be $\epsilon = \frac{L^2\epsilon_2}{2}$ to give an appropriate $N$ to be used.
Handling the case where $L$ is negative is similar, except you need to use the right hand part of \eqref{eq1} instead and make sure to use absolute values in appropriate places. I'll leave this part to you to finish.
