Prove that $b_k \to 1/a$ if $a_k \to a$ where $b_0 = 0$ and $b_k = 1/a_k$ for $k>0$

Question

Let $a_n$ where $n \in \mathbb {N}$ be a sequence of rational numbers converging to $a$. Suppose $a \neq 0$, for $k = 1, 2, ...$ let $$b_k=\begin{cases} 0 & \text{if}\;a_k=0\\\\\frac{1}{a_k} &\text{if}\;a_k \neq 0\end{cases}$$ Prove that $b_n$ converges to $\frac{1}{a}$.

I was studying real analysis and got stuck on this problem. Can you help me solve this problem or give me some hints?

Thanks

edit: is it possible to solve this in terms of Cauchy Sequence?

Answer 1

It is not necessary to use Cauchy sequences.

Let $\varepsilon = \frac{|a|}{2} > 0$ (because $a \neq 0)$. The sequence $(a_k)$ converges to $a$, so there exists $N \in \mathbb{N}$, such that for all $k \geq N$, $a-\varepsilon < a_k < a +\varepsilon$. By definition of $\varepsilon$, you get $a_k \neq 0$ for all $k \geq N$.

So, for all $k \geq N$, $b_k = \frac{1}{a_k}$. Taking the limit, you get immediately that $(b_k)$ converges to $\frac{1}{a}$.

Answer 2

Hint

Because $a\not=0$ (let's say $a>0$), show that $\exists n_0\in\mathbb{N}: \forall n\geq n_0\quad a_n>0$