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?
Best Answer
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