# 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 $$$$\label{eq:inv-limit} \left|{\frac{1}{a_n}-\frac{1}{L}}\right| = \left|{\frac{L-a_n}{a_nL}}\right| = \frac{\left|a_n- L\right|}{\left|a_nL\right|} < \frac{\varepsilon}{\left|a_n\right|\left|L\right|}\,.$$$$

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?

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

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$$0where $$L-\epsilon'$$ can be arbitrarily a large positive number since $$L$$ is positive and $$\epsilon'>0$$ is arbitrary

• Thanks! I'm having difficulty with the following, though: what happens when $L$ is negative? Oct 31, 2018 at 10:52
• 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 Oct 31, 2018 at 16:28

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 $$|\frac{1}{a_n}-\frac{1}{L}| \lt \epsilon$$.

• 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 $L-|L|/2 = |L|/2$. Oct 31, 2018 at 8:57