find the limit,if existsf(x)=ln(x-2)/(x-3), x->3
We need to find lim x--> 3 [ ln(x - 2)/(x - 3)]
lim x--> 3 [ ln(x - 2)/(x - 3)]
substituting x = 3 gives us the indeterminate form 0/0, so we can use l'Hopital's rule and substitute the numerator and denominator by their derivatives.
=> lim x--> 3[ (1/(x - 2))/1]
substitute x = 3
The required value of the limit is 1.
First, we'll substitute x by the value of accumulation point.
lim ln(x-2)/(x-3) = ln(3-2)/(3-3) = ln1/0 = 0/0
We've get an indetermination, so, we'll apply L'Hospital rule:
lim ln(x-2)/(x-3) = lim d[ln(x-2)]/dx/d(x-3)/dx
lim d[ln(x-2)]/dx/d(x-3)/dx = lim 1/(x-2)/1
We'll substitute again x by 2 and we'll get:
lim 1/(x-2)/1 = 1/(3-2) = 1
The limit of the function f(x), if x approaches to 3, is lim ln(x-2)/(x-3) = 1