Is $(x^3)/x$ really the same as $x^2$? Is $f(x)=\dfrac{x^3}{x}$ really the same as $x^2$? At $x=0$, $f(x)=\dfrac{x^3}x$ is undefined.  So then why can't I just say $f(x) = \dfrac{x^3}{x} = x^2$ therefore $f(x)=\dfrac{x^3}{x} $  at $x=0$ is $0$ and not undefined!
 A: On $$\underbrace{\mathbb R\setminus \{0\},}_{\mathbb R,\; \text{except 0}}\quad f(x) =x^2$$  
But that's pretty much what you concluded.
At $x=0,$ the original function is not defined $\frac{0}{0}$, (as you noted!) and so it is excluded from the domain.  
It is only when $x\neq 0$, that we can fully equate $$f(x) = \frac{x^3}{x} = x^2.$$
It is good that you realized the issue when $x=0$. In other contexts, multiplying and/or dividing by zero (unwittingly) often leads to incomplete, even fallacious solutions.
This is a very good example of removable discontinuity at $x=0$. 
If graphed, $f(x) = \frac{x^3}{x} = x^2$; the only difference in their graphs is that the original function will leave a hole, at the origin $(0, 0)$
A: Well, it depends on what exactly you mean with $\frac{x^3}{x}$. Now you may say, it should be obvious what is meant, but a closer inspection shows that it is not at all obvious.
The one interpretation that you have in mind is:

$\frac{f(x)}{g(x)}$ is the function $h(x)$ defined in all $x$ so that:

*

*the domain of $h$ is the intersection of the domain of $f$ and the support of $g$ (that is, $h$ is defined on all values where $f$ is defined and $g$ is both defined and nonzero), and


*on its domain, the function value of $h$ is the quotient of the function value of $f$ and the function value of $g$.

With this definition of the quotient, you are indeed right that the quotient is defined only for $x\ne 0$.
However, we could also note that both $x^3$ and $x$ are polynomials in $x$, and then use polynomial division as definition:

If $f(x)$ and $g(x)$ are polynomials, then $\frac{f(x)}{g(x)}$ is the unique polynomial $h(x)$ such that $f(x) = g(x)h(x)$, or undefined if no such polynomial exists.

Under that definition, indeed $\frac{x^3}{x}=x^2$, and you don't have to exclude any point, as the definition is not pointwise. On the other hand, using this definition, $\frac{x}{x^3}$ is not defined at all.
Now you can go further and use rational functions instead of polynomials, with appropriate definition of the division. Then you get both $\frac{x^3}{x}=x^2$ and $\frac{x}{x^3}$ defined.
A: When you are saying that $f(x)=\dfrac{x^3}{x}=x^2$ you are forgetting to add the condition that it only works when $x$ is not equal to zero.
A: Recall what it means to simplify a fraction:
$$\frac{ac}{bc} = \frac{\frac{ac}{c}}{\frac{bc}{c}} = \frac{a}{b}$$
When can the above occur? We divided by $c$, so it has to be $c \neq 0$. Therefore, in your example, we have to make a distinction. When $x \neq 0$, it's actually
$$f(x) = \frac{x^3}{x} = x^2$$
When $x = 0$ you cannot do that, as you would divide by zero. For this reason, $\operatorname{dom} f = \mathbb R \setminus \{0\}$. So yes, $f(x) = x^2$ for $x \in \operatorname{dom}f$.
Borrowing terms from measure theory, you would say that those two functions are equals almost everywhere.
A: Note that $f(x) = x^2\cdot x/x$, you can cancel out the last factor only if $x\ne 0$. In this case $f(x) = x^2$. That is $f(x) = x^2$ for $x\ne 0$. 
Now, it is true that
$$
\lim_{x\to 0} f(x) = \lim_{x\to 0} x^2\cdot x/x = \lim_{x\to 0} x^2 = 0
$$ 
I could cancel $x$ in this expression, because $x$ is actually never $0$ in the limit. So you have have that


*

*$f(x) = x^2$ for $x\ne 0$

*$\lim_{x\to 0} f(x) = 0$

*$f(0)$ is not determined

A: Well using rules of indices $X^a/X^b$ = $X^{a-b}$.
Therefore, $X^3/X$is the same as saying $X^{3-1}$therefore the answer is $X^2$.
However if you change the $X's$ into $0$ then the rules of indices then do not apply as you then have $0^3/0$.
And we know that any number divided by $0$ will give you a maths error and this is not allowed.
