It's like this.

I didn't have any problems to prove some of definitions of sequences like :

$\displaystyle lim_{n->+\infty}{(ax_n+by_n)}=ax+by$

Let

$\displaystyle lim_{n->+\infty}{x_n}=x$

and

$\displaystyle lim_{n->+\infty}{y_n}=y$

for any $\displaystyle \xi>0$. Starting from some index $\displaystyle n_1$ every member of sequence $\displaystyle (x_n)$ are in $\displaystyle \xi$-region of point x. Same, from some index $\displaystyle n_2$ every member of sequence $\displaystyle (y_n)$ are in $\displaystyle \xi$-region of point y.

If we put $\displaystyle n_0=max {\{n_1 ,n_2 \}}$ then for every $\displaystyle n >= n_0$ these inequality are true :

$\displaystyle |x_n-x|<\xi$ and $\displaystyle |y_n-y|<\xi$

so from inequality of triangle for $\displaystyle n >= n_0$ we have:

$\displaystyle |ax_n+by_n-(ax+by) |=|ax_n-ax+by_n-by|\le |a||x_n-x|+|b||y_n-y| \le (|a|+|b| )\xi$

Because $\displaystyle (|a|+|b|)$ is fix real number and $\displaystyle \xi$ any small number so it's $\displaystyle (|a|+|b| )\xi$ any small number so there we have that is:

$\displaystyle lim_{n->+\infty}{(ax_n+by_n)}=ax+by$

true.

This and some another I got it OK

But I can't,

or don't see how to prove (or show) that sum of two null-sequences are again null-sequence (and difference of two sequences) becose it's to obviously.