Question

$\underset{x\to 3}{\lim }\frac{x-3}{\left|x-3\right|},$ is equal to
​
(a) 1
(b) −1
(c) 0
(d) does not exist

Answer 1

(d) does not exist

$$\begin{gathered} \underset{x\to 3}{\lim }\frac{x-3}{\left|x-3\right|} \\ \mathrm{LHL}\mathrm{at}x=3: \\ \underset{x\to 3^{-}}{\lim }\frac{x-3}{-\left(x-3\right)}\left[\because \left|x-3\right|=-\left(x-3\right),\mathrm{when}x<3\right] \\ \Rightarrow -1 \\ \mathrm{RHL}\mathrm{at}x=3: \\ \underset{x\to 3^{+}}{\lim }\frac{x-3}{x-3}\left[\because \left|x-3\right|=x-3,\mathrm{when}x>3\right] \\ =1 \end{gathered}$$

LHL ≠ RHL
Therefore, limit does not exist.