Suppose that $x_n$ $\rightarrow$ + $\infty$and $y_n$ $\rightarrow$ a
True or false? If a>0, then $x_ny_n$ $\rightarrow$ + $\infty$
This is what I have:
Let $\epsilon$>0 then $n_0$ $\in$ $\mathbb{N}$ for |$y_n$-a|<$\epsilon$. Let $\epsilon$=a/2 then |$y_n$-a| < a/2 so, -a/2< $y_n$-a < a/2 add a to both sides and a/2< $y_n$ < 3a/2. Therefore, $y_n$>a/2 for n> $n_0$ if $y_n$> a/2 and $x_n$ $\rightarrow$ $\infty$. So, $x_ny_n$> $x_n$(a/2) which equals $x_ny_n$> +$\infty$ (a/2)= +$\infty$. Therefore the statement is true.
I was wondering if this would also work if $\epsilon$=a/4 with the proof:
Let $\epsilon$>0 then $n_0$ $\in$ $\mathbb{N}$ for |$y_n$-a|<$\epsilon$. Let $\epsilon$=a/4 then |$y_n$-a| < a/4 so, -a/4< $y_n$-a < a/4 add a to both sides and 3a/4< $y_n$ < 5a/4. Therefore, $y_n$>3a/4 for n> $n_0$ if $y_n$> 3a/4 and $x_n$ $\rightarrow$ $\infty$. So, $x_ny_n$> $x_n$(3a/4) which equals $x_ny_n$> +$\infty$ (3a/4)= +$\infty$. Therefore the statement is true.
or can $\epsilon$ only equal a/2?