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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?

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  • $\begingroup$ You also want $L \neq 0$. What is $L_a$? $\endgroup$ Oct 30, 2018 at 22:17
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    $\begingroup$ Hint: Can you show that $|a_n|$ is bounded? $\endgroup$
    – Daniel
    Oct 30, 2018 at 22:28

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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

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  • $\begingroup$ Thanks! I'm having difficulty with the following, though: what happens when $L$ is negative? $\endgroup$
    – Anakhand
    Oct 31, 2018 at 10:52
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    $\begingroup$ Any time! We can answer this question in two ways 1. if $a_n\to L$ when $L<0$ then $-a_n\to -L>0$ and ${1\over -a_n}=-{1\over a_n}\to -{1\over L}$ then ${1\over a_n}\to {1\over L}$ 2. as well as $L-\epsilon$ can be bounded from below when $L>0$, $L+\epsilon$ can be bounded from above when $L<0$. In both cases, $|L+\epsilon|$ and $|L-\epsilon|$ remain arbitrarily bounded $\endgroup$ Oct 31, 2018 at 16:28
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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 $|\frac{1}{a_n}-\frac{1}{L}| \lt \epsilon$.

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  • $\begingroup$ Thank you! There is just one step I'm not sure about: how can you go from $\left|a_n - L \right| < |L|/2$ to $\left|a_n\right| > |L|/2$? I guess one could reason something like $\left|a_n - L \right| < \frac{\left|L\right|}{2} \Rightarrow -\frac{\left|L\right|}{2} < a_n-L < \frac{\left|L\right|}{2} \Rightarrow a_n > L-\frac{\left|L\right|}{2}$, but I'm unsure how $L-|L|/2 = |L|/2$. $\endgroup$
    – Anakhand
    Oct 31, 2018 at 8:57

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