Analysis Limit Proofs

Question

I am confused on how I would prove the following:

Prove if $\lim_{x\to a}f(x)=L$ exists, $\lim_{x\to a}|f(x)|=|L|$. Also, show there is a function where $\lim_{x\to a}|f(x)|=|L|$ but $\lim_{x\to a}f(x)$ does not exist.

I really have no idea how to prove either. Any suggestions on how to begin?

Answer 1

Hints:

1. $\lvert x\rvert$ is a continuous function.
2. Think of a sequence with terms alternatively positive and negative : $(-1)^n u_n$.

Answer 2

Assuming that $\lim_{x\to a}f(x)=L$, you want to prove that $\lim_{x\to a}\bigl|f(x)\bigr|=|L|$. Let $\varepsilon>0$. You know that there is a $\delta>0$ such that $|x-a|<\delta\implies\bigl|f(x)-L\bigr|<\varepsilon$. But $\bigl||f(x)|-|L|\bigr|\leqslant\bigl|f(x)-L\bigr|$. Therefore, $|x-a|<\delta\implies\bigl||f(x)|-|L|\bigr|<\varepsilon$.

On the other hand, if$$\begin{array}{rccc}f\colon&\mathbb{R}&\longrightarrow&\mathbb{R}\\&x&\mapsto&\begin{cases}1&\text{ if }x\in\mathbb Q\\-1&\text{ if }x\notin\mathbb{Q},\end{cases}\end{array}$$then $\lim_{x\to0}\bigl|f(x)\bigr|=1$, but $\lim_{x\to0}f(x)$ does not exist.