Epsilon delta limits If
$\displaystyle\lim_{x\rightarrow 0} f(x)=0$
Then prove 
$\displaystyle\lim_{x\rightarrow 0} f(2x)=0$
I know that I need to use the epsilon-delta definition, so I have
$$
∀ ε>0\ ∃δ>0:∀x∈D(f) :0<|x|<δ⇒|f(x)|<ε
$$
How do I show this is true for $f(2x)$?

You have that $\lim_{x\to 0} f(x) = 0$, so the $\varepsilon$-$\delta$-definition is saying that
$$\forall \varepsilon >0, \ \exists \delta> 0 : \ \forall x\in D(f): \ 0<|x|<\delta \Rightarrow |f(x)|<\varepsilon.$$
Now, fix $\varepsilon>0$, and define $\delta_1 := \delta/2$, this $\delta_1$ works also for the previous limit. Try to see if it also Works for the second one.
