What is $\frac{x^2}{|x|}$ at $x=0$ Why $\frac{x^2}{|x|}$ is not defined on $x=0$. It should be $0$, since $\frac{x^2}{|x|}=|x|$, isn't?
 A: The expression
$$ \frac{x^2}{|x|}$$
is not defined when $x=0$.  However, we can find another expression that "does the same job" as $x^2/|x|$ and is defined at zero.  To do this, define the function
$$ f(x) := \frac{x^2}{|x|}, $$
which is not defined at zero.  On the other hand, the function
$$ g(x) := |x| $$
is defined for all real values of $x$, and $f(x) = g(x)$ whenever both functions are defined.  Thus it is reasonable to "extend" $f$ to all real numbers by setting $f(0) = 0$.  In this way, we can reasonably say that
$$ \frac{x^2}{|x|} = |x|, $$
which is zero when $x=0$.

At the suggestion of Henning Makholm, let us note that this $g$ is in no way unique.  I could just as easily have remarked that
$$ h(x) := \begin{cases} |x| & \text{if $x\ne 0$, and} \\
47 & \text{if $x = 0$} \\
\end{cases} $$
is equal to $f$ whenever both $h$ and $f$ are defined, therefore we could extend $f$ to all real numbers by setting $f(0) = 47$.  This approach is entirely reasonable, but somewhat unsatisfying.  We would like to extend $f$ to a function on $\mathbb{R}$ in the "nicest possible way."  What we mean by nicest possible way is nebulous, but in this context that probably means continuity (in other contexts it might mean smoothly, or up to a set of measure zero, or some other nonsensical mathematical gobbledygook).
That is, we would like to extend $f$ to a function on $\mathbb{R}$ so that the extended function is continuous.  The only way to to that is to set $f(0) = 0$.
A: $\dfrac{x^2}{|x|}\neq |x|$ when $x=0,\;$ since $\,\dfrac 1{|x|}\,$ is not defined at $x=0$
