Find the limit of function f(x) given by f(x)=ln(x-1)/(x-2), x->2?

First, we'll substitute x by the value of accumulation
point.

lim ln(x-1)/(x-2) = ln(2-1)/(2-2) = ln1/0 =
0/0

We've get an indetermination, so, we'll apply
L'Hospital rule:

lim ln(x-1)/(x-2) = lim
d[ln(x-1)]/dx/d(x-2)/dx

lim d[ln(x-1)]/dx/d(x-2)/dx = lim
1/(x-1)/1

We'll substitute again x by 2 and we'll
get:

lim 1/(x-1)/1 = 1/(2-1) =
1

The limit of the function f(x) is lim
ln(x-1)/(x-2) = 1.