Question

lim abs(x-3)/(x-3) as x approaches -3 from the right

Answer 1

Recall the definition of the absolute value:
\[|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x<0\end{cases}\]
Replacing \(x\) with \(x-3\) gives
\[|x-3|=\begin{cases}x-3&\text{for }x-3\ge0\\-(x-3)&\text{for }x-3<0\end{cases}\]
or
\[|x-3|=\begin{cases}x-3&\text{for }x\ge3\\-(x-3)&\text{for }x<3\end{cases}\]
Are you sure about the approaching value? Is \(x\to-3^+\)? Or do you mean \(x\to3^+\)?