## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to0^+}\Big(\frac{1}{x}-\frac{1}{|x|}\Big)=0$
$A=\lim\limits_{x\to0^+}\Big(\frac{1}{x}-\frac{1}{|x|}\Big)$ In this case, we consider $x\to0^+$, which means we only consider the values of $x\gt0$. While we know that, $$|x|=x\hspace{.5cm}for\hspace{.5cm}x\geq0$$ Therefore, $A=\lim\limits_{x\to0^+}\Big(\frac{1}{x}-\frac{1}{x}\Big)$ $A=\lim\limits_{x\to0^+}0$ $A=0$