Answer
To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x|\lt\delta\Rightarrow|f(x)|\lt\epsilon$$
Work Step by Step
$\lim_{x\to0}f(x)=0$ if $f(x)=2x$ for $x\lt 0$ and $f(x)=x/2$ for $x\ge0$
Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x|\lt\delta\Rightarrow|f(x)|\lt\epsilon$$
1) Solve the inequality $|f(x)|\lt\epsilon$ to find an open interval containing $x=0$ on which the inequality holds for all $x\ne0$
$$|f(x)|\lt\epsilon$$
However, since we have $2$ functions of $f(x)$, we need to divide into 2 cases:
*Case 1: For $x\lt0$, $f(x)=2x$. This also means $2x\lt0$, and $|2x|=-2x$
$$|2x|\lt\epsilon$$ $$-2x\lt\epsilon$$ $$x\gt-\frac{\epsilon}{2}$$
*Case 2: For $x\ge0$, $f(x)=x/2$. This also means $x/2\ge0$, and $|x/2|=x/2$
$$|\frac{x}{2}|\lt\epsilon$$ $$\frac{x}{2}\lt\epsilon$$ $$x\lt2\epsilon$$
Combining 2 cases, the open interval on which the inequality holds is $(-\epsilon/2,2\epsilon)$.
2) Find a value of $\delta\gt0$ that places the centered interval $(-\delta,\delta)$ inside the interval $(-\epsilon/2,2\epsilon)$
Take $\delta$ to be the distance from $0$ to the nearer endpoint of $(-\epsilon/2,2\epsilon)$. In other words, $\delta=\min[0-(-\epsilon/2),2\epsilon-0]=\min[\epsilon/2,2\epsilon]$
So if $\delta$ has this or any smaller positive value, then the constraint $0\lt|x|\lt\delta$ will automatically place $x$ between $-\epsilon/2$ and $2\epsilon$ so that $|f(x)|\lt\epsilon$.
That means for all $x$, $$0\lt|x|\lt\delta\Rightarrow|f(x)|\lt\epsilon$$
This completes our proof.
