The question:

Let f: D-> R and let c be an accumulation point of D. Suppose that lim_x->c f(x)=L.

Prove that lim_x->c |f(x)|=|L|.

A sketch of what I have:

Definition of limit and then

|f(x)-L|<$\displaystyle \varepsilon$

||f(x)|-|L|| $\displaystyle \leq$ |f(x)-L|<$\displaystyle \varepsilon$

But I'm sketchy on the reasons and if this is the right direction. Any ideas? Thanks.