Show that `f(x)=|x-3|` is not differentiable at `x=3.`

Explanation

`f(3)=0`

For differentiablility at `x=3`

`(L H D at x=3)=lim _{x -> 0^{-}} frac{f(x)-f(3)}{x-3}`

`=lim _{h -> 0} frac{f(3-h)-f(3)}{3-h-3}`

`=lim _{h -> 0} {|3-h-3|-0}/{3-h-3}`

`=lim _{h -> 0} h/{-h}=-1`

And

`(R H D text { at } x=3)=lim _{x -> 0^{+}} frac{f(x)-f(3)}{x-3} `

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