## University Calculus: Early Transcendentals (3rd Edition)

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$
$\lim_{x\to1}f(x)=2$ 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.