Answer
To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$
Work Step by Step
$\lim_{x\to1}f(x)=2$ if $f(x)=4-2x$ for $x\lt 1$ and $f(x)=6x-4$ for $x\ge1$
Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$
1) Solve the inequality $|f(x)-2|\lt\epsilon$ to find an open interval containing $x=1$ on which the inequality holds for all $x\ne1$
$$|f(x)-2|\lt\epsilon$$
However, since we have $2$ functions of $f(x)$, we need to divide into 2 cases:
*Case 1: For $x\lt1$, $f(x)=4-2x$
$$|4-2x-2|\lt\epsilon$$ $$|-2x+2|\lt\epsilon$$
Since $x\lt1$, $-2x\gt-2$ then $-2x+2\gt0$, meaning $|-2x+2|=-2x+2$ $$-2x+2\lt\epsilon$$ $$-2x\lt\epsilon-2$$ $$ x\gt\frac{\epsilon}{2}-1$$
*Case 1: For $x\ge1$, $f(x)=6x-4$
$$|6x-4-2|\lt\epsilon$$ $$|6x-6|\lt\epsilon$$
Since $x\ge1$, $6x-6\ge6-6=0$, meaning $|6x-6|=6x-6$ $$6x-6\lt\epsilon$$ $$6x\lt\epsilon+6$$ $$ x\lt\frac{\epsilon}{6}+1$$
Combining 2 cases, the open interval on which the inequality holds is $((\epsilon/2)-1,(\epsilon/6)+1)$.
2) Find a value of $\delta\gt0$ that places the centered interval $(1-\delta,1+\delta)$ inside the interval $((\epsilon/2)-1,(\epsilon/6)+1)$
Take $\delta$ to be the distance from $1$ to the nearer endpoint of $((\epsilon/2)-1,(\epsilon/6)+1)$. In other words, $\delta=\min[1-((\epsilon/2)-1),(\epsilon/6)+1-1]=\min[2-(\epsilon/2),\epsilon/6]$
So if $\delta$ has this or any smaller positive value, then the constraint $0\lt|x-1|\lt\delta$ will automatically place $x$ between $(\epsilon/2)-1$ and $(\epsilon/6)+1$ so that $|f(x)-2|\lt\epsilon$.
That means for all $x$, $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$
This completes our proof.
