17-07-2023
Problem: Let $U \subset \mathbb{R} $ be any open set and $f: U \to \mathbb{R} $ be a function. Let $a \in U$.
- If $\displaystyle \lim_{x \to a} f(x) = l$, then show that $\displaystyle \lim_{x \to a} \vert f(x) \vert = \vert l \vert $.
- Show that the converse need not be true.
Solution: Let $f: U \to \mathbb{R} $ be the given function.
- We are given that $\displaystyle \lim_{x \to a} f(x) = l$ and need to show that $\displaystyle \lim_{x \to a} \vert f(x) \vert = \vert l \vert $. Let $\varepsilon > 0$ be given. Since $\displaystyle \lim_{x \to a} f(x) = l$, so for the given $\varepsilon $ there exists $\delta >0$ such that \[ \vert x - a \vert \lt \delta \implies \vert f(x) - l \vert \lt \varepsilon . \] Now let $\vert x - a \vert \lt \delta$. Since \begin{align*} \vert \vert f(x) \vert - \vert l \vert \vert & \leq \vert f(x) - l \vert \lt \varepsilon . \end{align*} Therefore, $\displaystyle \lim_{x \to a} \vert f(x) \vert = \vert l \vert $.
-
The converse need not be true. For example, consider the function
\[
f(x) =
\begin{dcases}
1 - x, &\text{ if } x \leq 2;\\
x - 1, &\text{ if } x > 2.
\end{dcases}
\]
Note that $\displaystyle \lim_{x \to 2} f(x)$ does not exist.
On the other hand, $\displaystyle \lim_{x \to 2} \vert f(x) \vert $ exists and it is equal to $1$.
