Please dont laugh on my english if I make any mistake but im hungarian.
Now let's see the problem:
Proof if lim(an)=A and lim(bn)=B then lim(an×bn)=A×B
Thank for your help
The sequence $\displaystyle b_n$ is bounded: $\displaystyle \left( {\exists M} \right)\left[ {\left| {b_n } \right| < M} \right]$.
Then $\displaystyle \left| {a_n b_n - AB} \right| \leqslant \left| {a_n b_n - Ab_n } \right| + \left| {Ab_n - AB} \right| = \left| {b_n } \right|\left| {a_n - A} \right| + \left| A \right|\left| {b_n - B} \right|$
We have;
$\displaystyle \forall\ \epsilon>0\ \exists\ N\geq1\ \forall\ n\geq N \mid a_n-A \mid<\epsilon$
$\displaystyle \forall\ \epsilon>0\ \exists\ M\geq1\ \forall\ l\geq M \mid b_l-B\mid<\epsilon$
We want to proof;
$\displaystyle \forall\ \epsilon>0\ \exists\ K\geq1\ \forall\ k\geq K \mid a_kb_k-AB\mid<\epsilon$
Start the proof as follow;
$\displaystyle \mid a_kb_k-AB\mid \leq \mid a_kb_k-a_kB+a_kB-AB\mid \leq \mid a_kb_k-a_kB\mid + \mid a_kB-AB\mid $$\displaystyle = \mid a_k(b_k-B)\mid + \mid B(a_k-A)\mid = \mid a_k \mid \mid b_k-B \mid + \mid B \mid \mid a_k-A \mid$
Use that a convergence sequence is bounded to proof the remaining part.