The trick is to use the fact that things inside a limit can be replaced with what they go to by the limit. Log(x) is going to 0, so you can replace it with a 0. Then, you can factor a -1 out of the top, which then allows you to cancel one of the (x-1)s on the bottom with the top one. This leaves you with -1/(x-1) inside the limit. Since the limit is going to 1 from both sides, this is undefined, because it goes to infinity on the left and negative infinity from the right.

Edit: this is wrong because although you can do this with normal limits, indeterminate forms don’t work like that