**Proof**: Using triangle inequality, note that $\begin{align}|a_{n}b_{n}-ab| &= |a_{n}b_{n} -a_{n} b + a_{n} b -ab| \\ & \leq |a_{n} -a_{n}b| + |a_{n}b-ab| \\ &= |a_{n}| |b_{n}-b| + |b||a_{n}-a| \end{align}$since $(a_{n})$ converges by [[Convergent Real Sequence is Bounded]], $\exists A \geq 0$ such that $|a_{n}|$ for all $n$. Since $a_{n} \to a$ and $b_{n} \to b$, $\exists N_{1}, N_{2}$ such that for $n \geq \max(N_{1},N_{2})$ $|a_{n}- a| < \frac{\epsilon}{2(|b|+1)} \text{ and } |b_{n}-b|< \frac{\epsilon}{2A}$so $|a_{n}b_{n} - ab| < \epsilon$ which shows that $a_{n}b_{n} \to ab$ as claimed.