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

To prove the limit, prove that for every value of $\epsilon\gt0$, there exists a corresponding number of $\delta\gt0$ such that for all $x$: $$0-\delta\lt x\lt0\Rightarrow \Big|\frac{x}{|x|}-(-1)\Big|\lt\epsilon$$
$$\lim_{x\to0^-}\frac{x}{|x|}=-1$$ Given $\epsilon\gt0$. To prove the limit, we need to find $\delta\gt0$ such that for all $x$: $$0-\delta\lt x\lt0\Rightarrow \Big|\frac{x}{|x|}-(-1)\Big|\lt\epsilon$$ $$-\delta\lt x\lt0\Rightarrow \Big|\frac{x}{|x|}+1\Big|\lt\epsilon$$ First, we need to examine the inequality: $$\Big|\frac{x}{|x|}+1\Big|\lt\epsilon$$ $$-\epsilon\lt\frac{x}{|x|}+1\lt\epsilon$$ Since $-\delta\lt x\lt0$, here we examine only values of $x\lt0$, where $|x|=-x$. Therefore, $$-\epsilon\lt\frac{x}{-x}+1\lt\epsilon$$ $$-\epsilon\lt-1+1\lt\epsilon$$ $$-\epsilon\lt0\lt\epsilon$$ which is always right as we define $\epsilon\gt0$ In other words, for $x\lt0$, $\Big|\frac{x}{|x|}+1\Big|\lt\epsilon$ Take any arbitrary value of $\delta\gt0$, or $-\delta\lt0$, we would have $$-\delta\lt x\lt0\Rightarrow\Big|\frac{x}{|x|}+1\Big|\lt\epsilon$$ According to the defintion of a one-sided limit, this means $\lim_{x\to0^-}\frac{x}{|x|}=-1$