**Proof**: Note that $\left\lvert \frac{1}{b_{n}} - \frac{1}{b} \right\rvert = \left\lvert \frac{b_{n}-b}{b_{n}b} \right\rvert = \frac{|b_{n}-b|}{|b_{n}| |b|} $Since $b \neq 0$, $|b| \neq 0$. Since $b_{n} \to b$, $\exists N_{1}$ such that for all $n \geq N_{1}$, we have $|b_{n}-b|< \frac{|b|}{2}$so using triangle inequality $|b_{n}| \geq |b|- |b-b_{n}| > \frac{|b|}{2}$Take $\epsilon>0$. Since $b_{n} \to b$, $\exists N_{2}$ such that for all $n \geq N_{2}$ we have $|b_{n}-b|< \frac{\epsilon |b|^{2}}{2}$so for $n \geq N:= \max(N_{1},N_{2})$ we have $\left\lvert \frac{1}{b_{n}} - \frac{1}{b} \right\rvert = \frac{|b_{n}-b|}{|b_{n}| |b|} < \frac{\epsilon |b|^{2}}{2} \times \frac{2}{|b|^{2}} = \epsilon$so $\frac{1}{b_{n}} \to \frac{1}{b}$. So using product rule $a_{n}/b_{n}=a_{n}(1/b_{n})\to a(1/b)=a/b$.