$(b_n) \rightarrow b$ then $(1/b_n) \rightarrow 1/b$ proof. Let $(b_n)$ be a sequence that converges to $b$. Show that $(1/b_n)$ converges to $1/b$ when $b_n \neq 0$ for all $n$ and $b \neq 0$.
First see that
$$
|\frac{1}{b_n} - \frac{1}{b}| = |\frac{b - b_n}{b_nb}| = \frac{|b_n - b|}{|b_n||b|}
$$
Now from convergence of $(b_n)$ it follows that $\forall \epsilon \geq 0$ there is $N \in \mathbb{N}^{+}$ such that $|b_n - b| \leq \epsilon$ when $n \geq N$.
That implies that $\forall \epsilon \geq 0$ when $n \geq N$, it is true that 
$$
\frac{|b_n - b|}{|b_n||b|} \leq \frac{\epsilon}{|b_n||b|} = \epsilon'
$$
Clearly given any $\epsilon$ we have a positive $\epsilon'$, therefore for any
$\epsilon' \geq 0$ it follows that there is $N \in \mathbb{N}^{+}$ such that
$$
|\frac{1}{b_n} - \frac{1}{b}| \leq \epsilon'
$$
Hence $(1/b_n) \rightarrow 1/b$.
Correction based on yousef magableh answer:
I didn't notice that my $\epsilon'$ was defined in terms of $n$. To avoid that suppose that $b > 0$. Let's pick an $\epsilon = 2b$. Hence that will exists $N_1 \in \mathbb{N}^{+}$ that guarantees that:
$$
|b_n-b| \leq 2b \rightarrow -3b \leq -b \leq b_n \leq 3b \rightarrow |b_n| \leq 3b
$$
Therefore, for any positive $\epsilon$, there is a positive $\epsilon'$ such that:
$$
\frac{|b_n - b|}{|b_n||b|} \leq \frac{\epsilon}{|b_n||b|} = \frac{\epsilon}{3b|b|}  = \epsilon'
$$
holds for $n \geq N_1$.
If $b < 0$ just pick $\epsilon = -2b$.
Is it correct? If so, what would you do in order to improve it? Thanks.

No!. The value of N should not depend on $ {b_n}  $.
Hint: Try to bound $ \frac{1}{|b_n|}  $ (above) by bounding $ {|b_n|}  $ (below) using the fact that $ {b_n}  $ converges. Let epsilon = $ \frac {|b|}{2} $ in the definition of convergence for $ {b_n}  $ . And then you will have a bound on $ {|b_n|}  $ .
See:proof (1/bn) -> (1/b) from abbott Understanding Analysis..
