TITLE: Why do reciprocal functions work for undefined values?
QUESTION [11 upvotes]: I have a certain problem: When you have a function $f(x)$ and it can have some undefined values, the reciprocal function does some things I don't understand. For example: 
i) $$f(x)=\tan(x) $$
or another one:
ii)
$$ f(x)=\frac {5}{x-3}  $$
Clearly you can see that for ii) $x=3$ will be undefined and for i) $\pi/2, 3\pi/2,\ldots$
When I now work with the reciprocal functions my problem appears : 
i) $$(f(x))^{-1}=\frac{1}{\tan x}=\cot x$$
ii) $$(f(x))^{-1}=\frac{1}{\frac {5}{x-3}}$$
Coming up to my questions:
i) The function of $\cot x$ is $0$ at those $x$-values where $\tan x$ is undefined. Why is $1/\text{undefined}=0$ and not undefined too?
ii) The same here; $f(x)$ is undefined at $x=3$, but depending on which graphic calculator you use it defines  $$(f(3))^{-1}=0 $$ or as a discontinuity.
*What happens at those "magical" points when $f(x)=\text{undefined}$ and $(f(x))^{-1}= 1/\text{undefined?}$ 
What I got so far:

"If y = f (x) = 0 for some value of x, then 1/f (x) is undefined.
  There is a jump or discontinuity in its graph for this value of x.
  This means that, as f (x) gets close to 0, 1/f (x) will become very
  large in value. Equally, if there is a jump or discontinuity in the
  graph of y = f (x) for some value of x, then y = 1/f (x) = 0 for that
  value of x." 
  That's a definition I have from "Jenny Olive: Math a
   student's survival guide"

and of course : Why is cot(x)=0 instead of undefined
and my own thesis: 

Since you can convert $$ y= \frac {1}{\frac {5}{x-3}}=1\cdot\frac{x-3}{5}$$ the value for x=3 is defined
Same for $\cot(x) = \cos(x) / \sin(x)$ , where $x= \pi/2$ is defined.

Is my thesis working or am I hurting mathematics at this point?

REPLY [3 votes]: The premise of the question isn't true. The reciprocal of an undefined expression is still an undefined expression.
For your first example:

$$(f(x))^{-1}=\frac{1}{\tan x}=\cot x$$
...
The function of $\cot x$ is $0$ at those $x$-values where $\tan x$ is undefined. Why is $1/\text{undefined}=0$ and not undefined too?

The function $\cot x$ is $\frac{\cos x}{\sin x}$, not $\frac{1}{\tan x}$. The expression $\frac{1}{\tan x}$ is undefined when $\tan x$ is undefined, but this doesn't affect $\cot x$, because $\cot x$ and $\frac{1}{\tan x}$ are different.
For your second example:

$$ f(x)=\frac {5}{x-3}  $$
$$(f(x))^{-1}=\frac{1}{\frac {5}{x-3}}$$
The same here; $f(x)$ is undefined at $x=3$, but depending on which graphic calculator you use it defines  $$(f(3))^{-1}=0 $$ or as a discontinuity.

Well, regardless of what your graphic calculator says, $(f(3))^{-1}$ is undefined, because $f(3)$ is undefined.

and my own thesis:

Since you can convert $$ y= \frac {1}{\frac {5}{x-3}}=1\cdot\frac{x-3}{5}$$ the value for x=3 is defined
Same for $\cot(x) = \cos(x) / \sin(x)$ , where $x= \pi/2$ is defined.

Is my thesis working or am I hurting mathematics at this point?

It's not correct to simplify $\frac {1}{\frac {5}{x-3}}$ to $\frac{x-3}{5}$, because the denominator of $\frac {1}{\frac {5}{x-3}}$ is undefined when $x = 3$.

It may be worth noting that "undefined" isn't a mathematical value. The word "undefined" is an adjective, and it means "not having a definition". This is why mathematicians always write "is undefined", never "= undefined".
Some definitions are conditional, meaning that they only "work" under certain conditions. For example, the definition of division over the real numbers is:

If $x$ and $y$ are real numbers, and $y \ne 0$, then $\frac{x}{y}$ is that real number such that $\frac{x}{y} \cdot y = x$.

This definition tells us what, say, $\frac23$ is, but it does not give a definition for $\frac10$, so the expression $\frac10$ still does not have a definition.
The definition also doesn't give a result when either $x$ or $y$ is not defined as being a real number. That's why a fraction is undefined whenever its numerator or denominator is undefined.