Sequences of positive numbers and limits

1. Jul 31, 2005

iNCREDiBLE

Let (x(n)) and (y(n)) be sequences of positive numbers such that lim(x(n)/y(n)) = 0.
If lim(x(n)) = +∞, then lim(y(n)) = +∞
If (y(n)) is bounded, then lim(x(n)) = 0

To me this is self-evident. But HOW can it be proved?

2. Jul 31, 2005

Maxos

applications of definitions:

for any $$\epsilon > 0$$ there's an $$n_\epsilon :$$ if $$n > n_\epsilon$$ then $$|\frac {x_n} {y_n}| < \epsilon$$ and for any $$M > 0$$ there's an $$n_M :$$ if $$n > n_M$$ then $$|x_n| > M$$

then for any $$M > 0$$ there's a k := max {$$n_M; n_\epsilon$$} : if $$n > k$$ then $$|\frac {x_n} {y_n}| < \epsilon$$ and $$|x_n|> M$$ then $$\epsilon|y_n|>|x_n|> M$$ then $$|y_n|> \frac {M} {\epsilon} > M$$

CVD

Last edited: Jul 31, 2005
3. Jul 31, 2005

Maxos

The other is possibly simpler:

$$\exists M > 0 : \forall n \in \mathbb{N}, |y_n| < M$$
$$\forall \epsilon > 0, \exists n_{\epsilon} : n > n_{\epsilon} \Rightarrow |\frac {x_n} {y_n}| < \epsilon$$
$$\Rightarrow \forall \delta > 0, \exists n_{\delta} : n > n_{\delta} \Rightarrow \frac {|x_n|} {\delta} < |y_n| < M$$
$$\Rightarrow |x_n| < \delta M, \delta := \frac {\epsilon} {M}$$
$$\Rightarrow \exists k := max(n_\epsilon; n_\delta) : n > k \Rightarrow |x_n| < \epsilon$$

CVD

Last edited: Jul 31, 2005