limit with absolute value in denominator
By Andrew Adams •
I know that there are two scenarios where you use $(4-v)$ when $v\geq 4$ and $-(4-v)$ when $v<4$. But what is the answer and why/how?
$$\lim_{v\to 4^+}\frac{4-v}{|4-v|}$$
$\endgroup$ 42 Answers
$\begingroup$The notation $v \rightarrow 4^+$ implies that we consider $v > 4$.
$$\lim_{v \rightarrow 4^+} \frac{4-v}{|4-v|}=\lim_{v \rightarrow 4^+} \frac{4-v}{v-4}=-1$$
$\endgroup$ $\begingroup$$v \to 4^+$ means that
$v=4+ |h|$ with $h\to 0$.
the limit becomes
$$\lim_{h\to 0}\frac {-|h|}{|h|}=-1$$
$\endgroup$ 2
