Why does $|x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon$ imply $|f(x)|>|f(x_0)|-\epsilon$ for continuous f?

Question

Let $f$ be continuous and $f(x_0) \neq 0$. Let $\epsilon:=\frac{|f(x_0)|}{2}$.

Because f is continuous there exists a $\delta>0$ so that for every $x$ with $|x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon$.

Why does this imply that $|f(x)|>|f(x_0)|-\epsilon$?

Answer 1

Note that, by the triangle inequality, $$|f(x_0)|=|f(x)+f(x_0)-f(x)|\leq|f(x)|+|f(x_0)-f(x)|$$ Rearranging, we have $|f(x_0)-f(x)|\geq|f(x_0)|-|f(x)|,$ and thus $$\varepsilon>|f(x)-f(x_0)|=|f(x_0)-f(x)|\geq|f(x_0)|-|f(x)|.$$ Adding $|f(x)|-\varepsilon$ to both sides gives the desired result.

Answer 2

This is a consequence of reverse triangle inequality which states that for any real numbers $x$ and $y$: $$\left\vert \vert x\vert - \vert y \vert \right\vert \le \vert x - y \vert$$

Apply that to $f(x), f(x_0)$ you get: $$\vert f(x_0) \vert - \vert f(x) \vert \le \left\vert f(x_0) - f(x) \right\vert <\epsilon$$

Hence the requested result by adding $\vert f(x) \vert - \epsilon$ on both sides of the inequality.