Suppose the sequences of real numbers $a_{n}$ and $b_{n}$ converge to $a$ and $b$, respectively.
Then the sequence $(a_{n}b_{n})_{n=m}^{\infty}$ converges to $ab$.
MY ATTEMPT
To start with, let us notice that \begin{align*} |a_{n}b_{n} - ab| & = |a_{n}b_{n} - a_{n}b + a_{n}b - ab|\\\\ & = |a_{n}(b_{n} - b) + b(a_{n} - a)|\\\\ & \leq |a_{n}||b_{n} - b| + |b||a_{n} - a| \end{align*}
Since $a_{n}$ converges, it is bounded. That is to say, $|a_{n}| \leq A$.
Hence we can rewrite the given relation as \begin{align*} |a_{n}b_{n} - ab| \leq A|b_{n} - b| + |b||a_{n} - a| \end{align*}
Moreover, based on the convergence of $a_{n}$, for every $\varepsilon/2|b| > 0$, there exists a natural number $N\geq m$ such that \begin{align*} |a_{n} - a| \leq \frac{\varepsilon}{2|b|} \end{align*} whenever $n \geq N$.
Similarly, based on the convergence of $b_{n}$, for every $\varepsilon/2A > 0$, there exists a natural number $M\geq m$ such that \begin{align*} |b_{n} - b| \leq \frac{\varepsilon}{2A} \end{align*} whenever $n\geq M$. Finally, for every $\varepsilon > 0$, there is a natural $n_{0} = \max\{N,M\}$ such that \begin{align*} |a_{n}b_{n} - ab| \leq A|b_{n}-b| + |b||a_{n} - a| \leq A\times \frac{\varepsilon}{2A} + |b|\times\frac{\varepsilon}{2|b|} = \varepsilon \end{align*} whenever $n \geq n_{0}$, and we are done.
Could someone please double-check my reasoning? Any other solution would be greatly appreciated.