Suppose that the $\lim_{x\rightarrow c}{|f(x)|}$ exists. Give an example to show that $\lim_{x\rightarrow c}{f(x)}$ may not exist.

Question

Suppose that the $\lim_{x\rightarrow c}{|f(x)|}$ exists. Give an example to show that $\lim_{x\rightarrow c}{f(x)}$ may not exist.

I had done a similar problem using sequences rather than using functions where {$x_n$} = (-1)$^n$ was the correct sequence, but since that is not a function. I can not use the same idea for this problem.

Should I try to solve this using the Delta-Epsilon Definition or is there something else I am missing?

Answer 1

Something rather similar to your idea (which worked for sequences). Take $f$ defined on $\mathbb{R}$ by $$f(x) = \begin{cases} 1& \text{ if } x\in\mathbb{Q}\\ -1& \text{ if } x\in\mathbb{R}\setminus\mathbb{Q} \end{cases}$$ $\lvert f\rvert$ is constant, equal to $1$; so continuous everywhere, But $f$ is discontinuous everywhere.$^{(\dagger)}$

Note that you can replace $\mathbb{Q}$ by any set $A$ dense in $\mathbb{R}$ such that $\mathbb{R}\setminus A$ is dense as well.

$(\dagger)$ Indeed, for any $a\in\mathbb{R}$ and any $\delta>0$, the interval $(a-\delta, a+\delta)$ contains both rationals and irrationals, so $f$ will take both value $1$ and $-1$ in $(a-\delta, a+\delta)$. Now, take e.g. $\varepsilon = \frac{1}{2}$ to contradict continuity at $a$.

Answer 2

I had done a similar problem using sequences rather than using functions where {$x_n$} = (-1)$^n$ was the correct sequence, but since that is not a function. I can not use the same idea for this problem.

This trick divides the natural numbers into the even and odd numbers, producing the desired oscillating behavior, no longer present after taking the absolute value.

Hint: can you divide the real numbers into two subsets, again mapping $f$ to $\pm 1$ depending on the subset $x$ is in, to obtain a similar effect?

$$f : \mathbb{R} \to \mathbb{R} : x \mapsto \left\{ \begin{array}{rl} 1 & \quad x \in A \\ -1 & \quad x \in \mathbb{R} \setminus A \end{array}\right.$$

Answer 3

Define $$ f(x)=\begin{cases} x+1 & \text{if $-1\le x<0$} \\[4px] x-1 & \text{if $0<x\le 1$} \end{cases} $$ and consider the limits of $f(x)$ and $|f(x)|$ at $0$: \begin{align} \lim_{x\to0^-}f(x)&=\\ \lim_{x\to0^+}f(x)&=\\ \lim_{x\to0}|f(x)|&= \end{align}

Answer 4

Take $f(x):=\operatorname{sign}(\cos(\frac{1}{x}))$ for $x>0$ and $f(x)=1$ otherwise. For $x_n:=\frac{1}{n\pi}$ you get $\operatorname{sign}(\cos(x_n))=(-1)^{n+1}$, which doesn't converge, but $|f(x)|=1$ for all $x$ and thus the limit exists in this case.

Answer 5

Intuition: the absolute value does away with distinctions in sign. So in our search for counterexamples, we should look for functions where a sign difference breaks continuity, but the absolute value is continuous.

So how about the sign function, $$f(x) = \operatorname{sgn}(x) = \begin{cases} 1, & x \geq 0 \\ -1, & x < 0 \end{cases}$$ with $c=0$? We have $\lim_{x\to 0}|f(x)| = \lim_{x\to 0}1 = 1$ but $\lim_{x\to 0^-} f(x) = -1$ and $\lim_{x\to 0^+}f(x) = 1$, so $f$ is not continuous at $0$.