Question

Evaluate: $\underset{x\to 3}{lim}\frac{|x-3|}{x-3}$

Answer 1

As $x\to 3^{-},x-3<0$
$\therefore |x-3|=-(x-3)$
$\therefore \underset{x\to 3^{-}}{lim}\frac{|x-3|}{x-3}=\underset{x\to 3^{-}}{lim}\frac{-(x-3)}{x-3}=-1$
As $x\to 3^{+},(x-3)>0\therefore |x-3|=x-3$
$\underset{x\to 3^{+}}{lim}\frac{|x-3|}{x-3}=\underset{x\to 3^{+}}{lim}\frac{x-3}{x-3}=1$
Clearly L.H.S$\ne$R.H.S
Hence, the limit does not exist.