## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{v\to4^+}\frac{4-v}{|4-v|}=-1$$
$$A=\lim\limits_{v\to4^+}\frac{4-v}{|4-v|}$$ We see that $|4-v|=(4-v)$ if $(4-v)\ge0$ or $v\le4$ and $|4-v|=-(4-v)$ if $(4-v)\lt0$ or $v\gt4$ In this case, since $v\to4^+$, we only consider the values of $v\gt4$. Therefore, $|4-v|=-(4-v)$ So, $$A=\lim\limits_{v\to4^+}\frac{4-v}{-(4-v)}$$$$A=\lim\limits_{v\to4^+}\frac{1}{-1}$$$$A=\lim\limits_{v\to4^+}(-1)$$$$A=-1$$