edit: all of the $x$ in my proof and scratch work were supposed to be $x_0$... woops. I just deleted the confusing part in the beginning because I'm too lazy to change every $x$ into $x_0$ lol
All right, after my previous question, someone told me to try proving that $f(x)=1/x$ is continuous. I'm going to attempt this on just the positive side. I will again have scratch work before the $\epsilon-\delta$ proof. Here are the variables/constants I will use.
$c>0$ is an arbitrary constant. It will be the fixed point around which we will set our boundary with some $\delta>0$. $x>0$ is an arbitrary constant in $(c-\delta,c+\delta)$ so that the resulting $f(x)$ is in the range $(f(c)-\epsilon,f(c)+\epsilon)$. For a more aesthetically pleasing work, I will use $\beta=1/c$. $1/c$ will appear a lot.
$$Scratchwork:$$
$0<|x-c|<\delta$
$|{1\over x}-{1\over c}|<\epsilon\Rightarrow-\epsilon<{1\over x}-{1\over c}<\epsilon\Rightarrow\beta-\epsilon<{1\over x}<\beta+\epsilon$
$\Rightarrow{1\over \beta+\epsilon}<x<{1\over \beta-\epsilon}\Rightarrow{1\over \beta+\epsilon}-c<x-c<{1\over \beta-\epsilon}-c$
Now I will set two conditions: $x>c$ or $x<c$ (but $x>0$ and $c>0$).
If $x>c$, then $x-c>0$ so
$$x-c=|x-c|<{1\over \beta-\epsilon}-c$$
If $x<c$, then $x-c<0$ so $x-c=-|x-c|>{1\over \beta+\epsilon}-c$
$$\Rightarrow|x-c|<c-{1\over \beta+\epsilon}$$
So we can let $\delta$ equal to one of the two last expressions, depending on the relative sizes of $x$ and $c$.
/end of scratch work
$$Proof:$$
$\delta=min${$c-{1\over \beta+\epsilon},{1\over \beta-\epsilon}-c$}
Here, I define $y:={1\over x}-{1\over c}$.
This will have two parts: one for $x>c$ and another for $x<c$.
$1.$ If $x<c$, then ${1\over x}>{1\over c}$.
${1\over x}=y+{1\over c}=y+\beta\Rightarrow x={1\over y+\beta}\Rightarrow x-c={1\over y+\beta}-c$
Because $x<c$, $x-c$ is negative and thus $x-c=-|x-c|$ so $|x-c|=c-{1\over y+\beta}<c-{1\over \beta+\epsilon}$ (from scratch work)
$\Rightarrow c-{1\over y+\beta}<c-{1\over \beta+\epsilon}\Rightarrow {1\over y+\beta}>{1\over \beta+\epsilon}\Rightarrow y+\beta<\beta+\epsilon$
$\Rightarrow y<\epsilon\Rightarrow {1\over x}-{1\over c}<\epsilon$
Remember how $x<c$ so ${1\over x}>{1\over c}$? Therefore, ${1\over x}-{1\over c}=|{1\over x}-{1\over c}|<\epsilon$
$\Rightarrow |{1\over x}-{1\over c}|<\epsilon$
Now for $x>c$...
- If $x>c$, then ${1\over x}<{1\over c}$
${1\over x}-{1\over c}=y\Rightarrow {1\over x}=y+{1\over c}=y+\beta$
$\Rightarrow x={1\over y+\beta}\Rightarrow x-c={1\over y+\beta}-c$
$x>c$ so $x-c=|x-c|={1\over y+\beta}-c<{1\over \beta-\epsilon}-c$ (from scratchwork)
$\Rightarrow {1\over y+\beta}<{1\over \beta-\epsilon}\Rightarrow y+\beta>\beta-\epsilon\Rightarrow y>-\epsilon$
$\Rightarrow {1\over x}-{1\over c}>-\epsilon$ but ${1\over x}-{1\over c}=-|{1\over x}-{1\over c}|>-\epsilon$ (because ${1\over x}<{1\over c}$)
$\Rightarrow |{1\over x}-{1\over c}|<\epsilon$
Thus, if $0<|x-c|<\delta$ and $\delta=min${$c-{1\over \beta+\epsilon},{1\over \beta-\epsilon}-c$}, then $|{1\over x}-{1\over c}|<\epsilon$, demonstrating that ${1\over x}$ is continuous on $(0,\infty)$.
I am aware of the proof using the fact that $|{1\over x}-{1\over c}| = |{x-c\over xc}|$; but that's not the one I first came up with when I first tried the proof. The one I've written out just now is the first. Please let me know if you spot any errors, and thanks in advance!