Counterexample for $\lim \limits_{x \to c} \left\lvert f(x) \right\rvert = \left\lvert L \right\rvert$ then $\lim \limits_{x \to c} f(x) = L$

Question

I know that the converse of this is true. I was looking for some counterexamples proving this statement false.

Answer 1

Consider:

$$f(x)=\begin{cases} 1 \text{ if } x\in \mathbb{Q}\\ -1 \text{ if } x\not\in \mathbb{Q}\end{cases}$$

Then $\lim\limits_{x\to 1} |f(x)|=1$ but $\lim\limits_{x\to 1}f(x)$ doesn't exist.

Answer 2

The Heaviside function can be defined as $H(x) =1$ for $x\ge 0$ and $H(x)=-1$ for $x<0$. Clearly $\lim_{x\to 0}|H(x)|=1$, but $\lim_{x\to 0}H(x)$ does not exist since the left and right sided limits are of opposite sign.

In fact, this counter example falls into a class of functions with jump discontinuities with right and left sided limits of opposite sign.

Answer 3

Take $f(x) = -x$ and consider say $x \to 3$ then

$$ \lim_{x \to 3} |f(x)| =3 $$ but this does not mean that

$$ \lim_{x \to 3 }f(x) = 3 $$ Other nice examples are given above.

Answer 4

look at $$f(x) = \frac{x}{|x|}.$$ we have $\lim_{x\to 0} |f(x)| = 1$ and $\lim_{x \to 0} f(x)$ is undefined because $\lim_{x \to 0+}f(x) = 1$ and $\lim_{x \to 0-}f(x) = -1.$