show using the $\epsilon - \delta$ criterion: $ \quad f(x)= \frac{x^2}{|x|} \quad \lim_{x\rightarrow 0} f(x) = 0$ Show using the $\epsilon - \delta $ criterion: 
$f: \mathbb R \backslash \{0\} \rightarrow$ with $f(x)= \frac{x^2}{|x|}$ has the limit: $\lim_{x\rightarrow 0} f(x) = 0$.
Question: Is my proof correct and valid?
Let $x_0=0.0001, \quad L=0, \quad x \in [-\infty, \infty] \backslash \{0\}, \quad\delta < \frac{\epsilon}{x^2}, \quad \epsilon > 0$
\begin{align}
|f(x)-0| = \left| \frac{x^2}{|x|} - 0 \right| = \left| \frac{x^2}{|x|} \right| =  \frac{|x^2|}{|x|} = \frac{x^2}{|x|} < x^2 < x^2 \delta = x^2  \frac{\epsilon}{x^2} = \epsilon,  \forall x \in D: 0 < |x - 0.0001| < \delta  
\end{align}
 A: In this kind of questions I like to separate the function in two functions. 
For $x>0$ you have $f1(x)=x^2/x=x$, 
and for $x<0$ you have $f2(x)=x^2/(-x)=-x$.
Then you observe $f(0+d)=f1(0+d)=d, d>0$.
And also $f(0-d)=f2(0-d)=-(-d)=d, d>0$.

From https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit#Precise_statement_for_real_valued_functions. You have c=0 and L=0. 
You have to prove that for every $e>0$ there exists $d$ such that for all x, $0<|x|<d$ means $|f(x)|<e$.
for $x>0$ $f(x)=x$ so $|f(x)|=|x|=x$. 
for $x<0$ $f(x)=-x$ so $|f(x)|=|-x|=-x$ (not $-x$ as $x$ is negative). 
Take e=d:
for $x>0$  if $0<|x|<d$ then $|f(x)|=x<d=e$
for $x<0$  if $0<|x|<d$ then $|f(x)|=-x<d=e$
A: There are plenty of mistakes


*

*you set $x_0=0.00001$ but this is completely irrelevant and not needed.

*$\delta$ can't be dependent on $x$ so $\delta<\varepsilon/x^2$ make no sense

*Instead of $x\in[-\infty,\infty]\backslash\{0\}$ it should be $x\in(-\delta,\delta)\backslash\{0\}$. (Note that $[-\infty,\infty]$ is problematic because you can't contain $-\infty$ and $\infty$ it should be something like $(-\infty,\infty)$ but that's not relevant anyway).
Let me help you with the formalism you begin that way:
Let $\varepsilon>0$ there exists $\delta = ?$ (only depending on $\varepsilon$!!) such that for every $x\in(-\delta,\delta)\backslash\{0\}$ we have that
$$|f(x)-0| = \left| \frac{x^2}{|x|} - 0 \right| = \left| \frac{x^2}{|x|} \right| =  \frac{|x^2|}{|x|} = \frac{x^2}{|x|} $$
If $x\in (0,\delta)$ then $$|f(x)-0|=\frac{x^2}{|x|} = \frac{x^2}{x}=x<\delta$$
If $x\in (-\delta,0)$ then $$|f(x)-0|=\frac{x^2}{|x|}=-\frac{x^2}{x}=-x<\delta$$
we conclude that for every $x\in(-\delta,\delta)\backslash\{0\}$ we have that $|f(x)-0|<\delta$
Now taking $\delta = (???)$ we have that $|f(x)-0|<\delta$ implies that $|f(x)-0|<\varepsilon$. 
So can you tell what $\delta$ equals to?
A: Consider an $\epsilon>0$. We want to find such $\delta>0$ that $0<|x|<\delta$ implies $|x^2/|x||=|x|<\epsilon$. Obviously $\delta=\epsilon$ works.
