The arguments around limits are addressed towards the end (under "Update 8/12/2018"):
> > If 0/0 = 0 then lim_(x -> 0) sin(x) / x = sin(0) / 0 = 0, but by L’Hospitals’ Rule lim_(x -> 0) sin(x) / x = lim_(x -> 0) cos(x) / 1 = 1. So we have 0 = 1.
> This was a really clever one. The issue is that the counterargument assumes that if the limit exists and f(0) is defined, then lim_(x -> 0) f(x) = f(0). This isn’t always true: take a continuous function and add a point discontinuity. The limit of sin(x) / x is not sin(0) / 0, because sin(x) / x is discontinuous at 0. For the unextended division it’s because sin(0) / 0 is undefined, while for our extended division it’s a point discontinuity. Funnily enough if we instead picked x/0 = 1 then sin(x) / x would be continuous everywhere.
Similar examples can be constructed for any regular function which is discontinuous (e.g. Heaviside step function).
> > If 0/0 = 0 then lim_(x -> 0) sin(x) / x = sin(0) / 0 = 0, but by L’Hospitals’ Rule lim_(x -> 0) sin(x) / x = lim_(x -> 0) cos(x) / 1 = 1. So we have 0 = 1.
> This was a really clever one. The issue is that the counterargument assumes that if the limit exists and f(0) is defined, then lim_(x -> 0) f(x) = f(0). This isn’t always true: take a continuous function and add a point discontinuity. The limit of sin(x) / x is not sin(0) / 0, because sin(x) / x is discontinuous at 0. For the unextended division it’s because sin(0) / 0 is undefined, while for our extended division it’s a point discontinuity. Funnily enough if we instead picked x/0 = 1 then sin(x) / x would be continuous everywhere.
Similar examples can be constructed for any regular function which is discontinuous (e.g. Heaviside step function).