Question

limits?How does this equal -1?

Answer 1

plugging in x=3 gives you the indeterminate 0/0

use l'hopital's rule

Answer 2

\[|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x<0\end{cases}\]
which means
\[|3-x|=\begin{cases}3-x&\text{for }3-x\ge0\\-(3-x)&\text{for }3-x<0\end{cases}\]
or
\[|3-x|=\begin{cases}3-x&\text{for }x\le3\\-(3-x)&\text{for }x>3\end{cases}\]
Since \(x\to3^+\), we know that \(x>3\), which means
\[\large\lim_{x\to3^+}\frac{3-x}{|3-x|}=\lim_{x\to3^+}\frac{3-x}{-(3-x)}=\lim_{x\to3^+}(-1)=-1\]