# Prove: $\lim_{x\to a}\lvert f(x) \rvert=L \rightarrow\Big[\lim_{x\to a} f(x) =L \lor \lim_{x\to a} f(x)=-L \Big]$, where $L \geq 0$

I am trying to prove the following statement:

$$\displaystyle \lim_{x\to a}\lvert f(x) \rvert=L \rightarrow\Big[\displaystyle \lim_{x\to a} f(x) =L \lor \displaystyle \lim_{x\to a} f(x)=-L \Big]$$, where $$L \geq 0$$.

We are therefore trying to arrive at either of these two final statements:

$$1)\displaystyle \lim_{x\to a} f(x) =L \iff \forall \epsilon \gt 0\ \exists\delta \gt 0 \text{ s.t. } \forall x \in \mathbb R \big [ 0\lt \lvert x -a \rvert \lt \delta \rightarrow\Big\lvert f(x) - L \Big\rvert \lt \epsilon \big ]$$

$$2) \displaystyle \lim_{x\to a} f(x) =-L \iff \forall \epsilon \gt 0\ \exists\delta \gt 0 \text{ s.t. } \forall x \in \mathbb R \big [ 0\lt \lvert x -a \rvert \lt \delta \rightarrow\Big\lvert f(x) - (-L) \Big\rvert \lt \epsilon \big ]$$

I will show my attempt but I have my doubts on the approach, which I will explain subsequently.

By assumption we have:

$$\displaystyle \lim_{x\to a}\lvert f(x) \rvert=L \iff \forall \epsilon \gt 0\ \exists\delta \gt 0 \text{ s.t. } \forall x \in \mathbb R \big [ 0\lt \lvert x -a \rvert \lt \delta \rightarrow\Big\lvert \lvert f(x) \rvert - L \Big\rvert \lt \epsilon \big ]$$

Choose an arbitrary error term and distance away from $$a$$: $$\epsilon'$$ and $$\delta_{\epsilon'}$$.

Then we have:

$$\forall x \in \mathbb R \big [ 0\lt \lvert x -a \rvert \lt \delta_{\epsilon'} \rightarrow\Big\lvert \lvert f(x) \rvert - L \Big\rvert \lt \epsilon' \big ]$$

Suppose $$f(x) \lt 0$$. Then we have $$\lvert f(x) \rvert = -f(x)$$. Thus:

$$\Big\lvert \lvert f(x) \rvert - L \Big\rvert = \Big\lvert -f(x) - L \Big\rvert=\Big \lvert (-1)(f(x)+L)\Big \rvert=\Big \lvert f(x)+L\Big \rvert=\Big \lvert f(x)-(-L)\Big \rvert \lt \epsilon'$$

Suppose, instead, $$f(x) \geq 0$$. Then we have $$\lvert f(x) \rvert = f(x)$$. Thus:

$$\Big\lvert \lvert f(x) \rvert - L \Big\rvert =\Big\lvert f(x) -L\Big\rvert \lt \epsilon'$$

My concern for the above proof is that I have not accounted for the fact that the $$x$$ used in the $$f(x) \geq 0$$ (or in the $$f(x) \lt 0$$) argument is taken from a pool of other possible values...all of which satisfy $$0 \lt \lvert x-a \rvert \lt \delta_{\epsilon'}$$. Just because $$f(x) \geq 0$$ (or $$f(x) \lt 0$$) does not necessarily mean that some other $$x'$$ (also from the pool of values satisfying $$0 \lt \lvert x'-a \rvert \lt \delta_{\epsilon'}$$) satisfies the same expression of $$f(x') \geq 0$$ (or $$f(x') \lt 0$$).

Because of this, I do not think my approach works. Any suggestions?

• The statement is false, unless you assume that limit of f at a exists Commented Apr 21, 2021 at 1:49
• @ArcticChar ahhh, that would make sense!
– S.C.
Commented Apr 21, 2021 at 1:59

The assertion is false: the standard counterexample is

$$f(x) = \begin{cases} 1 & \text{ if } x\ge 0, \\ -1 & \text{ if } x<0.\end{cases}.$$

Then $$|f(x)|=1$$ for all $$x$$, but $$f$$ has no limit at $$0$$.

The assertion holds if you assume that limit at $$a$$ exists: but then the proof is quite simple: if $$\lim_{x\to a} f(x) = M$$, then (see here)

$$\lim_{x\to a} |f(x)| = |M|,$$ which implies $$|M| = L$$, or $$M = \pm L$$.

• My attempt at playing around with this proof was a result of trying to build some intuition about statements of the form $\displaystyle \lim_{x \to a} \lvert f(x) \rvert = \infty$. Would the proof be solvable if a 3rd disjunction term was added? i.e. $...\lor \text { the limit does not exist }$ (I'm assuming so).
– S.C.
Commented Apr 21, 2021 at 2:10
• @S.Cramer Yes, that would do. Commented Apr 21, 2021 at 2:13