Elementary doubts on limits and absolute value functions I have elementary questions about the definition of the absolute value functions and their limits.
It should be remarked that I have also searched and read previous posts of Mathstackexchange. However, these did not help me with the questions to be presented here. Owing to the fact that there are few references on the subject, I think that these questions, and their possible answers, may be of great interest to those that subscribe to mathstacksexchange.
With that said, let me recall first the definition of the absolute value of a real function. It is well-known that we may define a modulus function as follows:
$f(x)=|x|=\left\{ \begin{array}{rcl}
x & \mbox{if}
& x \geq 0 \\ -x & \mbox{if} & x < 0 .
\end{array}\right.$
In the light of the foregoing, I ask:
1.Suppose that we have a function $\psi (x)$ that shows the possibility to diverge to positive infinity when $\psi >0$ as $x\rightarrow x_{0}$ or diverge to negative infinity when $\psi <0$ as $x\rightarrow x_{0}$.  Here, $x_{0}$ is real a constant. On account of the definition expressed above, is it correct to write that
$\displaystyle\lim_{x\rightarrow x_{0}}\psi \rightarrow \pm \infty \Leftrightarrow |\psi|\rightarrow \infty \quad\mbox{in which}\left\{ \begin{array}{rcl}
\psi \rightarrow \infty & \mbox{if}
& \psi > 0 & \mbox{as} & x \rightarrow x_{0}\\ \psi \rightarrow -\infty & \mbox{if} & \psi < 0 & \mbox{as} & x \rightarrow x_{0}   \quad ?
\end{array}\right.$


*Second, is there a more formal way to express this from the standpoint of mathematics?


*In addition, may you suggest references on the limits of absolute value functions that diverge to infinity?
Ps: It is worth emphasizing that I am not very well familiar with the correct notation of absolute value functions and limits. Hence, correct me if the above is not right.
 A: There are quite a few things that should be explained.
a) Don't write $\displaystyle\lim_{x\rightarrow x_{0}}\psi \rightarrow \pm \infty$. Write $\displaystyle\lim_{x\rightarrow x_{0}}\psi(x) = \pm \infty$. The limit is equal to some number, the function can go (tend/converge) to infinity. Of course sometimes the limit is a function itself and it can converge to something, but not here.
b) The statement $\displaystyle\lim_{x\rightarrow x_{0}}\psi(x) = \pm \infty$ means $\displaystyle\lim_{x\rightarrow x_{0}}\psi(x) = \infty$ or $\displaystyle\lim_{x\rightarrow x_{0}}\psi(x) = -\infty$. The statement $\displaystyle\lim_{x\rightarrow x_{0}}|\psi(x)| = \infty$ means something diferent. For example:

*

*If $f(x)=\frac 1x$ then $\lim_{x\to 0}f(x)$ doesn't exist but $\lim_{x\to 0}|f(x)|=\infty$.

*If $\displaystyle f(x)=\begin{cases}\frac 1x;&x\in\Bbb Q\setminus\{0\}\\ -\frac 1x;&x\notin\Bbb Q\end{cases}$ then $\lim_{x\to 0}f(x)$ doesn't exist (even $\lim_{x\to 0+}f(x)$ or $\lim_{x\to 0-}f(x)$) but $\lim_{x\to 0}|f(x)|=\infty$.
Therefore the answer to (1) is 'no'.

If 2. is concerned, you can sometimes write
$$\lim_{x\rightarrow x_{0}\pm}\psi(x) = \pm \infty \text{ or } \lim_{x\rightarrow x_{0}\pm}\psi(x) = \mp \infty$$ but if you want to precisely write what you described at the beginning of 1. you can use $\lim_{x\rightarrow x_{0}}|\psi(x)| = \infty$.
On the other hand, be careful with the case $\psi(x)=0$, that you didn't took into account. For example it's not clear what do you want to do the function like $f(x)=1/x$ for $x>0$ and $f(x)=0$ for $x<0$. It does satisfy the condition from the beginning of 1., but not the condition $\lim_{x\rightarrow x_{0}}|\psi(x)| = \infty$.
A: Question 1 : Yes. It is correct.
Question 2 : There may be other ways to write it, but your way is formal too.
Question 3 : There are many Examples:

*

*$1/x$ & $1/|x|$ at $x=0$


*$\tan(x)$ & $|\tan(x)|$ at $x=\pi/2$


*$1/(x-a)$ & $1/|x-a|$ at $x=a$


*$1/(e-e^x)$ & $1/|e-e^x|$ at $x=1$
