If $\lim_{x \to x_0} f(x) = L$, then $\lim_{x \to x_0} \lvert f(x)\rvert = \lvert L \rvert$.

Question

If $\lim_{x \to x0} f(x) = L$, then $\lim_{x \to x0} \lvert f(x)\rvert = \lvert L \rvert$.
I know this is true, because $\lvert f(x) \rvert - \lvert L \rvert <= \lvert f(x) - L \rvert < \epsilon$ but why is it bigger than minus Epsilon?

Answer 1

It follows from the fact that $$\vert \vert a \vert - \vert b \vert \vert \leq \vert a - b \vert$$ for all $a,b \in \mathbb{R}$. So if $\vert x - x_0 \vert < \delta$, then $$\vert \vert f(x) - \vert L \vert \vert \leq \vert f(x) - L \vert < \epsilon,$$ i.e. $\lim_{x \to x_0}\vert f(x) \vert = \vert L \vert$.

Answer 2

Just use the other side of the triangle inequality

$$\epsilon > \vert f(x) - L \vert = \vert L - f(x) \vert \geq \vert L \vert - \vert f(x) \vert \implies -\epsilon < \vert f(x) \vert - \vert L \vert$$