Prove that $\lim_{x\rightarrow 0}|x|=0$ Please check my proof 
There exist $\delta $such that $0<|x|<\delta $ imply $|x|< \epsilon $
Then choose $\delta =\epsilon $
for x is real number and $0<|x|<\delta \rightarrow |f(x)-0|=|0-0|< \delta =\epsilon $
Then limit is 0
 A: No, let $\varepsilon>0$ and $\delta=\varepsilon$.
if $|x|<\delta$, then,
$$|f(x)-0|=|x|<\delta=\varepsilon,$$
and the claim follow.
A: 
There exist $\delta$ such that $0<\lvert x\rvert<\delta$ imply $\lvert
 x\rvert < \epsilon$.

In such a type of proof for
$$
\lim_{x\to 0} \lvert x \rvert = 0
$$
you want to establish, that it is possible to have $\lvert x \rvert$ come arbitrary close to $0$ (here $0$ being the value of the limit), if $x$ can be chosen from a neighbourhood around $0$ (here $0$ being the argument where the limit is asked for).
The arbitrary closeness is formalized as a challenge for any positive $\epsilon$:
$$
\lvert x - 0 \rvert < \epsilon \quad (*)
$$
has to be achieved by being able to come up with a positive $\delta$ which confines $x$ to a neighbourhood around $0$:
$$
\lvert x - 0 \rvert < \delta \quad (**)
$$
and will imply that $(*)$ holds.
In this case it is easy, just choose $\delta = \epsilon$ and this $\delta$ will do the job. 
Then $(**)$ will lead to $(*)$.
A: Try this if you like.
Let $\epsilon>0$. We take $\delta=\epsilon$. So, we were able to find $\delta>0$ such that if assume $0<|x-0|<\delta$, then
$$\Big| |x|-0\Big|=|x|<\delta=\epsilon.$$
Then apply the definition of the limit to conclude that 
$$\lim_{x\to 0}|x|=0.$$
