About the limit of the absolute value of a function

Question

I'm trying to prove $\lim _{x\to x_0}f\left(x\right)=L$ then $\lim _{x\to x_0}\left|f\left(x\right)\right|=\left|L\right|$ (for $L\geq 0$ of course).

I proved it for $L = 0$ but struggled to prove it for $L > 0$. Any ideas?

Answer 1

An easy proof of the general statement is to use the reverse triangle inequality: $$\Big||f(x)|-|L|\Big|\leq |f(x)-L|$$ Let $\varepsilon>0$ be given. Since $f(x)\rightarrow L$ there exists $\delta>0$ such that $$0<|x-x_0|<\delta\Rightarrow |f(x)-L|<\varepsilon$$ The triangle inequality gives then that $$0<|x-x_0|<\delta\Rightarrow \Big||f(x)|-|L|\Big|<\varepsilon$$ which means that $|f(x)|\rightarrow |L|$. Alternatively, you could summon the squeeze theorem.