I'm stumped on this one.

Question

lim |x-1|sin(1/x-1)
x->1

Answer 1

as written, there's no problem. Just plug in 1 to get

|1-1| (sin 1/1-1) = 0.

However, I think you mean

|x-1|sin(1/(x-1))

again, no problem. Since sin is always less than 1, we have 0 * z, where z<1, so zero again.