How to prove that if $\lim(a_n)=a$ then $\lim(\frac{1}{a_n})=\frac{1}{a}$?

Question

How to prove that if $\lim(a_n)=a$ then $\lim(\frac{1}{a_n})=\frac{1}{a}$ ?

So far I have the following written down

$|a_n-a|<\epsilon$

$|\frac{a}{a_n}-\frac{1}{a}|=\frac{|a_n-a|}{a_na}$

and because $a_n$ must come within 9/10 of a, we have $a_n>0.9*a$

So we write: $<\frac{|a_n-a|}{\frac{9}{10}a^2}$

But now I am stuck and need assistance

Answer 1

This result is false if $a=0$

If $a_n \neq0$ and $a_n \rightarrow a\ne0 $, then the conclusion $\lim_{n \rightarrow \infty} \frac{1}{a_n}=\frac{1}{a}$ is true .

For, $a_n \rightarrow a $ implies $$(\forall \varepsilon>0,\exists N \in \Bbb{N}\;\text{and}\;\forall n \geq N): \vert a_n-a \vert < \varepsilon$$

Here $\vert a \vert >0$, so take in particular $\varepsilon=\frac{\vert a \vert}{2}$, so $$ \vert a_n-a \vert < \frac{\vert a \vert}{2}=\varepsilon, \forall n \geq N$$

Here, $$-\frac{\vert a \vert}{2}=-\varepsilon \leq -\vert a_n-a \vert \leq \vert a_n \vert-\vert a \vert,\forall n \geq N $$ $\Big[$Here, the second inequality follows from the fact $\Big\vert\; \vert a_n \vert -\vert a \vert\;\Big\vert \leq \vert a_n-a \vert \;\Big]$ and so $$\frac{1}{\vert a_n \vert}\leq \frac{2}{\vert a \vert}$$

Now, $$\Bigg\vert \frac{1}{a_n}-\frac{1}{a} \Bigg\vert= \Bigg\vert \frac{a-a_n}{a\cdot a_n} \Bigg\vert=\frac{\vert a-a_n \vert}{\vert aa_n \vert} \leq \frac{2}{\vert a \vert^2} \vert a-a_n \vert$$

Now try to make the RHS to be small and hence the result follows!