University Calculus: Early Transcendentals (3rd Edition)

To prove the limit, prove that for a $\epsilon\gt0$, there always exists a corresponding value of $\delta\gt0$ such that for all $x$: $$2\lt x\lt2+\delta\Rightarrow \Big|\frac{x-2}{|x-2|}-1\Big|\lt\epsilon$$
$$\lim_{x\to2^+}\frac{x-2}{|x-2|}=1$$ Given $\epsilon\gt0$. To prove the limit, we need to find $\delta\gt0$ such that for all $x$: $$2\lt x\lt2+\delta\Rightarrow \Big|\frac{x-2}{|x-2|}-1\Big|\lt\epsilon$$ First, we need to examine the inequality: $$\Big|\frac{x-2}{|x-2|}-1\Big|\lt\epsilon$$ $$-\epsilon\lt\frac{x-2}{|x-2|}-1\lt\epsilon$$ Since $2\lt x\lt2+\delta$, here we examine only values of $x\gt2$, where $|x-2|=x-2$. Therefore, $$-\epsilon\lt\frac{x-2}{x-2}-1\lt\epsilon$$ $$-\epsilon\lt1-1\lt\epsilon$$ $$-\epsilon\lt0\lt\epsilon$$ which is always right as we define $\epsilon\gt0$ In other words, for $x\gt2$, $\Big|\frac{x-2}{|x-2|}-1\Big|\lt\epsilon$ Take any arbitrary value of $\delta\gt0$, we would have $$2\lt x\lt2+\delta\Rightarrow\Big|\frac{x-2}{|x-2|}-1\Big|\lt\epsilon$$ According to the defintion of one-side limit, this means $\lim_{x\to2^+}\frac{x-2}{|x-2|}=1$