- #1

- 15

- 0

1/x = 1/x for x < 0

1/x = undefined for x = 0

1/x = 1/x for x > 0

δ(x) = 0 for x < 0

δ(x) = undefined for x = 0

δ(x) = 0 for x > 0

then for any part of 1/x that is defined, 0 is being added to it. Is it okay to ignore the fact that the function is being changed at x = 0 because it was already undefined?