real analysis - $left{frac{1}{f(x_n)}right}$ converges to $frac{1}{f(x)}$.

Let $f:\Bbb R \to \Bbb R$ be a continuous function. Let $\{x_n\}_{n=1}^\infty$ be a convergent sequence in $\Bbb R$ with $\lim \limits_{n\to\infty}x_n=x$ and $f(x)\ne 0$

I want to show that $\left\{\frac{1}{f(x_n)}\right\}$ converges to $\frac{1}{f(x)}$.

Now It would seem I have:

$$|f(x_n) - f(x)| \lt \epsilon$$
$$|f(x_n) - f(x)| \leq |f(x_n)| - |f(x)|$$

and now I don't get it, I would expect to get some relationship also less than epsilon, times by $-1$ to inverse the epsilon inequality and then find a way to get the ricipricals and that would reverse the epsilon inequality again giving me the result, but I can't see it.