Find the limit of function f(x) given by f(x)=ln(x-1)/(x-2), x->2?
We have to find the value of lim x-->2 [ ln (x - 1)/(x - 2)]
substituting x = 2, we get the indeterminate form 0/0; that permits the use of l'Hopital's Theorem and we can substitute the numerator and denominator with their derivatives
=> lim x-->2 [ (1/(x - 1))/1]
substitute x = 2
=> 1/(2 - 1)
The required value of lim x-->2 [ ln (x - 1)/(x - 2)] = 1.
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.