Convergence of Sequences Proof Let $(x_n)$ and $(y_n)$ be convergent sequences. Use the definition of convergence (no limit theorems!) to prove that the sequence $(3x_n2y_n)$ converges. 
I'm having trouble doing this using the definition of convergence and no limit theorems 
 A: Let:
$$x_{n} \mathop{\rightarrow}^{n \rightarrow \infty} x  \: \text{  and }  \: y_{n} \mathop{\rightarrow}^{n \rightarrow \infty} y$$
Let $\epsilon >0$.
The sequenses $x_n$ and $y_n$ are bounden (because they converge) so:
$$\exists M>0, \forall n \in \mathbb{N}:  |x_n|<M \text{ and } |y_n| <M$$
We have:
$$|6x_{n}y_{n}-6xy| = \\|6x_{n}y_{n}-6x_{n}y+6x_{n}y-6xy| \leq \\|6x_{n}y_{n}-6x_{n}y|+|6_{n}y-6xy| = 6|x_{n}||y_{n}-y|+|x_{n}-x|6|y| \\ \leq 6M|y_{n}-y|+6|x_{n}-x||y|$$
Can you proceed? Hint: you know you can get $|y_{n}-y|$ and $|x_{n}-x|$ arbitrary small.
A: Suppose $x_n\to x$ and $y_n \to y$. Since $x_n$ and $y_n$ are convergent, they are bounded; there exists $M_1,M_2 > 0$ such that $|x_n| < M_1$ and $|y_n| < M_2$ for all $n\in \Bbb N$. Furthermore, given $\epsilon > 0$, there exist indicies $N_1, N_2$ such that $|x_n - x| < \epsilon/(12M_2)$ for all $n\ge N_1$ and $|y_n - y| < \epsilon/(12M_1)$ for all $n \ge N_2$.  If $n \ge \max\{N_1,N_2\}$, then 
\begin{align}|(3x_n)(2y_n) - (3x)(2y)| &= 6|(x_n - x)y_n + x(y_n - y)|\\
& \le 6(|(x_n - x)||y_n| + |x||y_n - y|)\quad \text{by the triangle inequality}\\
& < 6\left(\frac{\epsilon}{12M_2}\cdot M_2 + M_1\cdot \frac{\epsilon}{12M_1}\right)\\
& = \frac{\epsilon}{2} + \frac{\epsilon}{2}\\
& = \epsilon.
\end{align}
Since $\epsilon$ was arbitrary, $(3x_n)(2y_n)$ converges (to $(3x)(2y) = 6xy$).
A: Suppose $x_n \to x$ and $y_n \to y$. Then $$\begin{align}\left|x_ny_n-yx \right| = \left|x_ny_n-yx +(x_ny-x_ny)\right| \\ \leq \left|x_ny_n-x_ny \right|+\left|x_ny-xy \right| \\ \leq |x_n||y_n-y|+|y||x_n-x|\end{align}$$ Since $(x_n)$ is a convergent sequence, it is bounded. We can then find some maximum value $M$ such that $|x_n| \leq M$. Hence $$|x_n||y_n-y|+|y||x_n-x|\leq M|y_n-y|+|y||x_n-x|$$ Now we can appeal to the definition of convergence. Given $\varepsilon>0$, there exists $N_1,N_2$ such that $$|y_n-y|< \frac{\varepsilon}{2M} \implies M|y_n-y|<\frac{\varepsilon}{2}$$ for all $n>N_1$ and $$|x_m-x|<\frac{\varepsilon}{2|y|}\implies |y||x_m-x|<\frac{\varepsilon}{2}$$ for all $m>N_2$ (don't forget to deal with the case of $y=0$, it is always true that $|0||x_m-x|<\frac{\varepsilon}{2}$). Let $N = \max\{N_1,N_2\}$ which will satisfy both inequalities for all $n>N$. Hence,  $$\varepsilon = \frac{\varepsilon}{2}+\frac{\varepsilon}{2}>M|y_n-y|+|y||x_m-x|>|x_ny_n-xy|$$ So a product of convergent sequences is convergent. It still remains to be shown that a constant times a convergent sequence is convergent. 
