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`f(x)= {{:((|x-4|)/(2(x-4)), if x ne 4),(0,if x = 4):}` at `x = 4` is.

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LHL of f(x) at x=4 is
`underset(xto4^(+))limf(x)=underset(hto0)limf(4-h)`
`=underset(hto0)lim(|4-h-4|)/(4-h-4)=underset(hto0)lim(|-h|)/(-h)`
`=underset(hto0)limh/-h=underset(hto0)lim-1`
`=-1`
`RHL " of " f(x) " at "x=4" is"`
`underset(xto4^(+))limf(x)=underset(hto0)limf(4+h)`
`=underset(hto0)lim(|4+h-4|)/(4+h-4)`
`=underset(hto0)lim(|h|)/(h)=underset(hto0)limh/h=underset(hto0)lim1`
`=1`

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