Both \((a_n-a)\) and \((b_n-b)\) are null and hence so is
\begin{equation*}
(a_n-a+b_n-b) = (a_n+b_n-(a+b))\text{.}
\end{equation*}
That means \(a_n+b_n \to a+b\text{.}\) Also,
\begin{align*}
(a_nb_n) & = (a_n-a)(b_n-b) + (ab_n) + (ba_n) -
(ab)\\
& \to 0 +ab + ba -ab = ab
\end{align*}
Since \(b \neq 0\text{,}\) there are \(0\lt m \lt M\text{,}\) such that for all \(n\) sufficiently large, \(m \lt |b_n|
\lt M\text{.}\) And so,
\begin{equation*}
\frac{|b-b_n|}{bM} \lt \frac{|b-b_n|}{|bb_n|} \lt
\frac{|b-b_n|}{bm}
\end{equation*}
\begin{equation*}
\left| \frac{1}{b_n} - \frac{1}{b} \right| =
\frac{|b-b_n|}{bb_n} \to 0.
\end{equation*}