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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?

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

1 Answer 1

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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 $M = \pm L$.

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  • $\begingroup$ 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). $\endgroup$
    – S.C.
    Commented Apr 21, 2021 at 2:10
  • $\begingroup$ @S.Cramer Yes, that would do. $\endgroup$ Commented Apr 21, 2021 at 2:13

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