real analysis - $|x_n| to infty implies |f(x_n)| to infty$

Let $f:\mathbb{R} \to \mathbb{R}$ a continuous function.

Show that

1) $\lim \limits_{x\to +\infty} |f(x)|=\lim \limits_{x\to -\infty}|f(x)|=+\infty$

implies

2) $|x_n|\to \infty \implies |f(x_n)|\to \infty$.

I know that $f$ is continuous iff $x_n \to a \implies f(x_n)\to f(a)$.

But, can I use

$\lim|f(x_n)|= |f(\lim(x_n))|=\lim\limits_{x\to \infty} |f(x)|=+\infty$

for infinite cases so directly?

If the solution is not this way, could you help me by giving a hint how to do it?

Answer

The first condition means

$$\forall M \quad \exists \bar x : \forall x\, |x|>\bar x \quad |f(x)|>M$$

and from here the second follows, indeed we have

$$\forall \bar x \quad \exists \bar n : \forall n>\bar n \quad |x_n|> \bar x$$

that is

$$\forall M \quad \exists \bar n : \forall n>\bar n \quad |f(x_n)|>M$$