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

Answer

N/a

Explanation

We have, `f(x)= {{:((|x-4|)/(2(x-4)), if x ne 4),(0,if x = 4):}` at `x = 4`.
At `x = 4, LHL = underset(xrarr4^(-))(lim) (|x-4|)/(2(x-4))`
`= underset(hrarr0)(lim)(|4-h-4|)/(2[(h-4)-4]) = underset(hrarr0)(lim) (|0-h|)/((8-2h-8))`
` = underset(hrarr0)(lim) (h)/(-2h) = (-1)/(2)` and `f(4) = 0 ne LHL`
So, `f(x)` is discontinuous at `x = 4`.