$\lim_{x\to x_0}|f(x)|=|\lim_{x\to x_0}f(x)|$

Question

Suppose that $f:D\to \mathbb{R}$ has a limit at an accumulation point $x_0\in D$. Prove that $|f|:D\to \mathbb{R}$ $x_0$ is an accumulation point and that $\lim_{x\to x_0}|f(x)|=|\lim_{x\to x_0}f(x)|$

Answer 1

Let $y:=\lim_{x\to x_0}f(x)$ which exists by assumption. For $x\in D$ observe that $$||f(x)|-|y||\leq |f(x)-y|$$ by the reverse triangle inequality. Since the r.h.s. approaches $0$ when $x\to x_0$, so does the l.h.s. and we arrive at the desired conclusion, formally: Fix some $\varepsilon>0$ and choose $\delta>0$ s.t. $|f(x)-y|<\varepsilon$ whenever $|x-x_0|<\delta$, $x\in D$. Then for the $x\in D$ satisfying the latter condition we get: $||f(x)|-|y||\leq\varepsilon$.