If a function `f` is defined as `f(x)={(|x-4|)/(x-4)\ \ \ ,\ \ \ x!=4 0\ \ \ ,\ \ \ x=4` Show that `f` is everywhere continuous except at `x=4` .

Explanation

Given,

`f(x)=(|x-4|)/(x-4)`

So,

`lim_(x->4-)f(x)`

`=>lim_(x->4-)(|x-4|)/(x-4)
`

`=>lim_(h->4)(|(4-4)-4|)/((4-4)-4)`

`=>lim_(h->4)(|(4-4)-4|)/((4-4)-4)`

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